Leetcode总结之Tree

package Tree;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Deque;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Stack;
class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
    }
}
class TreeLinkNode {
    int val;
    TreeLinkNode left, right, next;

    TreeLinkNode(int x) {
        val = x;
    }
}
class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int x) {
        val = x;
    }
}
public class TreeProblems {
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        TreeLinkNode a1 = new TreeLinkNode(1);
        TreeLinkNode a2 = new TreeLinkNode(2);
        TreeLinkNode a3 = new TreeLinkNode(3);
        TreeLinkNode a4 = new TreeLinkNode(4);
        TreeLinkNode a5 = new TreeLinkNode(5);
        TreeLinkNode a6 = new TreeLinkNode(6);
        TreeLinkNode a7 = new TreeLinkNode(7);
        a1.left = a2;
        a1.right = a3;
        a2.left = a4;
        a2.right = a5;
        a3.left = a6;
        a3.right = a7;
        System.out.println(a5.next.val);
        TreeNode aaa1 = new TreeNode(4);
        TreeNode aaa2 = new TreeNode(1);
        TreeNode aaa3 = new TreeNode(2);
        TreeNode aaa4 = new TreeNode(3);
        aaa1.left = aaa2;
        aaa2.left = aaa3;
        aaa3.left = aaa4;
        /*
         * 9 5 13 1 7 10
         */
        TreeNode aa1 = new TreeNode(5);
        TreeNode aa2 = new TreeNode(4);
        TreeNode aa3 = new TreeNode(8);
        TreeNode aa4 = new TreeNode(11);
        TreeNode aa5 = new TreeNode(13);
        TreeNode aa6 = new TreeNode(4);
        TreeNode aa7 = new TreeNode(7);
        TreeNode aa8 = new TreeNode(2);
        TreeNode aa9 = new TreeNode(1);
        aa1.left = aa2;
        aa1.right = aa3;
        aa2.left = aa4;
        aa3.left = aa5;
        aa3.right = aa6;
        aa4.left = aa7;
        aa4.right = aa8;
        aa6.right = aa9;
    }
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//树的问题很多都是dfs的问题，首先要学会preorder，inorder，postorder三种dfs算法的两种实现方式。
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
    /*
     * 94. Binary Tree Inorder Traversal 
     * 2015.11.21 By Mingyang
     * 跟preorder一样的dfs，不过是每次pop出来的时候记在小本本上，也就是记在list里面 中序遍历迭代解法
     * 用栈先把根节点的所有左孩子都添加到栈内， 然后输出栈顶元素，再处理栈顶元素的右子树
     */
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        if (root == null)
            return list;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = root;
        while (!stack.isEmpty() || p != null) {
            if (p != null) {
                stack.push(p);
                p = p.left;
            } else {
                TreeNode t = stack.pop();
                list.add(t.val);
                p = t.right;
            }
        }
        return list;
    }
    public List<Integer> inorderTraversal1(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        dfs(root, res);
        return res;
    }
    public void dfs(TreeNode root, List<Integer> res) {
        if (root == null)
            return;
        dfs(root.left, res);
        res.add(root.val);
        dfs(root.right, res);
    }
    /*
     * 144. Binary Tree Preorder Traversal 
     * 2015.11.21 By Mingyang
     * 典型的dfs的做法，每次push进去一个值的时候就记在小本本上，其实也就是记在list里面
     */
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        if (root == null)
            return list;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = root;
        while (!stack.isEmpty() || p != null) {
            if (p != null) {
                stack.push(p);
                list.add(p.val);
                p = p.left;
            } else {
                TreeNode t = stack.pop();
                p = t.right;
            }
        }
        return list;
    }
    public List<Integer> preorderTraversal1(TreeNode root) {
        List<Integer> res=new ArrayList<Integer>();
        dfs(res,root);
        return res;
    }
    public void dfs(List<Integer> res,TreeNode root){
        if(root==null)
          return;
        res.add(root.val);
        dfs(res,root.left);
        dfs(res,root.right);
    }
    /*
     * 145. Binary Tree Postorder Traversal 
     * 2015.11.21 by Mingyang
     * 左右中，反过来就是中左右，所以就相当于inorder反过来，稍稍改进一下就好
     */
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        if (root == null)
            return list;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = root;
        while (!stack.isEmpty() || p != null) {
            if (p != null) {
                stack.push(p);
                list.add(p.val);
                p = p.right;
            } else {
                TreeNode t = stack.pop();
                p = t.left;
            }
        }
        Collections.reverse(list);
        return list;
    }
    public List<Integer> postorderTraversal1(TreeNode root) {
        List<Integer> res=new ArrayList<Integer>();
        dfs(root,res);
        return res;
    }
    public void dfs1(TreeNode root,List<Integer> res){
        if(root==null)
         return;
        dfs1(root.left,res);
        dfs1(root.right,res);
        res.add(root.val);
    }
    /*
     * 95. Unique Binary Search Trees II
     * 2015.11.23 By Mingyang
     * 分成两个函数来写，另外一个函数写满起点到终点
     */
    public List<TreeNode> generateTrees(int n) {
        return generateTrees(1, n);
    }
    public List<TreeNode> generateTrees(int start, int end) {
        List<TreeNode> list = new LinkedList<TreeNode>();
        if (start > end) {
            list.add(null);
            return list;
        }
        for (int i = start; i <= end; i++) {
            List<TreeNode> lefts = generateTrees(start, i - 1);//以i作为根节点，左子树由[1,i-1]构成
            List<TreeNode> rights = generateTrees(i + 1, end);//右子树由[i+1, n]构成
            for (TreeNode left : lefts) {//左边和右边分别返回了一个list，每个值就是一个TreeNode，就代表一种可能性
                for (TreeNode right : rights) {
                    TreeNode node = new TreeNode(i);
                    node.left = left;
                    node.right = right;
                    list.add(node);//存储所有可能行
                }
            }
        }
        return list;
    }
    /*
     * 145. Binary Tree Postorder Traversal
     * 上面的变体，给你一个preorder的array，看看是不是BST的preorder 
     * 2015.1.24 by Mingyang
     */
    public static boolean canRepresentBST(int pre[], int n) {
        // Create an empty stack
        Stack<Integer> s = new Stack<Integer>();
        // Initialize current root as minimum possible
        // value
        int root = Integer.MIN_VALUE;
        // Traverse given array
        for (int i = 0; i < n; i++) {
            // If we find a node who is on right side
            // and smaller than root, return false
            if (pre[i] < root) {
                return false;
            }
            // If pre[i] is in right subtree of stack top,
            // Keep removing items smaller than pre[i]
            // and make the last removed item as new
            // root.
            while (!s.empty() && s.peek() < pre[i]) {
                root = s.peek();
                s.pop();
            }
            // At this point either stack is empty or
            // pre[i] is smaller than root, push pre[i]
            s.push(pre[i]);
        }
        return true;
    }
    /*
     * 156.Binary Tree Upside Down
     * 2016-3-26 by Mingyang
     * For example:
     *Given a binary tree {1,2,3,4,5},
        1
       / 
      2 3
     / 
    4   5
    *return the root of the binary tree [4,5,2,#,#,3,1].
        4
       / 
      5   2
     / 
    3   1
    *先去处理最左下角的分支，因为他就是未来的root，然后再一层一层的上来
     */
    public TreeNode UpsideDownBinaryTree(TreeNode root) {  
        if (root == null)  
            return null;  
        TreeNode parent = root, left = root.left, right = root.right;  
        if (left != null) {  
            TreeNode ret = UpsideDownBinaryTree(left);  
            left.left = right;  
            left.right = parent;  
            return ret;  
        }  
        return root;  
    } 
    /*
     * 272.Closest Binary Search Tree Value II 
     * 2016-3-27 By Mingyang
     * 二叉搜索树的中序遍历就是顺序输出二叉搜索树，所以我们只要中序遍历二叉搜索树，同时维护一个大小为K的队列，
     * 前K个数直接加入队列，之后每来一个新的数（较大的数），如果该数和目标的差，相比于队头的数离目标的差来说，
     * 更小，则将队头拿出来，将新数加入队列。如果该数的差更大，则直接退出并返回这个队列，因为后面的数更大，差值也只会更大。
     * @return true if result is already found.
     */
    public List<Integer> closestKValues(TreeNode root, double target, int k) {
        LinkedList<Integer> list = new LinkedList<Integer>();
        closestKValuesHelper(list, root, target, k);
        return list;
    }
    private boolean closestKValuesHelper(LinkedList<Integer> list,TreeNode root, double target, int k) {
        if (root == null) {
            return false;
        }
        if (closestKValuesHelper(list, root.left, target, k)) {
            return true;
        }
        if (list.size() == k) {
            if (Math.abs(list.getFirst() - target) < Math.abs(root.val - target)) {
                return true;
            } else {
                list.removeFirst();
            }
        }
        list.addLast(root.val);
        return closestKValuesHelper(list, root.right, target, k);
    }
    /*
     *     270    Closest Binary Search Tree Value 
     * Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.
     *  2016-3-26 by Mingyang
     *  其实很简单，就是在dfs的过程中不断的缩小范围，没到一个点就开始去update最小的距离
     */
    public int closestValue3(TreeNode root, double target) {
        int closest = root.val;
        while (root != null) {
            // 如果该节点的离目标更近，则更新到目前为止的最近值
            closest = Math.abs(closest - target) < Math.abs(root.val - target) ? closest: root.val;
            // 二叉搜索
            root = target < root.val ? root.left : root.right;
        }
        return closest;
    }
    /*
     * 265.Count Univalue Subtrees 
     * 2016-3-26 by Mingyang
     * Given a binary tree, count the number of uni-value subtrees.
     * A Uni-value subtree means all nodes of the subtree have the same value.
     * 这里是自底向上的模式，
     */
    private int count = 0;
    public int countUnivalSubtrees(TreeNode root) {
        unival(root);
        return count;
    }
    private boolean unival(TreeNode root) {
        if (root == null)
            return true;
        if (root.left == null && root.right == null) {
            count++;
            return true;
        }
        boolean left = unival(root.left);//自底向上第一步，走到最底的那一层
        boolean right = unival(root.right);
        if (left && right && (root.left == null || root.left.val == root.val)&& (root.right == null || root.right.val == root.val)) {
            count++;
            return true;
        }
        return false;
    }
    /*
     * 285.Inorder Successor in BST 
     * 11.21 By Mingyang
     * Case1: Node has right subtree: Go deep to leftmost node in right subtree or find min in right subtree
     * Case2:No right subtree:1.go back to parent from left, parent is successor,and parent is unvisited
     * 2.from right, parent already visited, we need on more further
     * ---Go to the nearest ancestor for which given node would be in left subtree
     */
     public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
            Stack<TreeNode> stk = new Stack<TreeNode>();
            TreeNode curr = root;
            // 找到目标节点并把路径上的节点压入栈,也就是首先找到这个节点，大概时间是logn
            while(curr != p){
                stk.push(curr);
                if(curr.val > p.val){
                    curr = curr.left;
                } else {
                    curr = curr.right;
                }
            }
            // 如果目标节点有右节点，则找到其右节点的最左边的节点，就是下一个数
            if(curr.right != null){
                curr = curr.right;
                while(curr.left != null){
                    curr = curr.left;
                }
                return curr;
            } else {
            // 如果没有右节点，pop栈找到第一个比目标节点大的节点。因为如果目标没有右节点，那么就看他从左边还是右边回到父亲，如果从左边回到父亲，父亲就是下一个
            //如果不是，go to the nearest ancestor for which given node would be in left subtree
                while(!stk.isEmpty() && stk.peek().val < curr.val){
                    stk.pop();
                }
                // 如果栈都pop空了还没有比目标节点大的，说明没有更大的了
                return stk.isEmpty() ? null : stk.pop();
            }
        }
    /*
     * 255.Verify Preorder Sequence in Binary Search Tree 
     * Given an array of numbers,
     * verify whether it is the correct preorder traversal sequence of a binary
     * search tree. 
     * 11.21 By Mingyang
     * Time complexity O(n^2), space complexity O(1). 
     */
      public boolean verifyPreorder(int[] preorder) {
            if (preorder == null || preorder.length <= 1) {
                return true;
            }         
            return dfs(preorder, 0, preorder.length - 1);
        }     
        private boolean dfs(int[] preorder, int low, int hi) {
            if (low >= hi) {
                return true;
            }      
            int root = preorder[low];
            int i = low + 1;
            while (i <= hi && preorder[i] < root) {
                i++;
            }         
            int j = i;
            while (j <= hi && preorder[j] > root) {
                j++;
            }        
            if (j <= hi) {
                return false;
            }        
            return dfs(preorder, low + 1, i - 1) && dfs(preorder, i, hi);
        }
    /*
     * 103. Binary Tree Zigzag Level Order Traversal 
     * 11.21 By Mingyang
     */
    public static ArrayList<ArrayList<Integer>> zigzagLevelOrder2(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if (root == null)
            return res;
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        int num = 0;
        boolean reverse = false;// a flag
        while (!queue.isEmpty()) {
            num = queue.size();
            ArrayList<Integer> levelres = new ArrayList<Integer>();
            for (int i = 0; i < num; i++) {
                TreeNode node = queue.poll();
                levelres.add(node.val);
                if (node.left != null)
                    queue.add(node.left);
                if (node.right != null)
                    queue.add(node.right);
            }
            if (reverse) {
                Collections.reverse(levelres);
                reverse = false;
            } else {
                reverse = true;
            }
            res.add(levelres);
        }
        return res;
    }
    /*
     * 107. Binary Tree Level Order Traversal II 11.21 By Mingyang I的解答加一行：
     * Collections.reverse(results);就好了
     */
    /*
     * 102. Binary Tree Level Order Traversal 
     * 11.21 by Mingyang 最基本的queue的做法
     */
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        ArrayList<Integer> result = new ArrayList<Integer>();
        if (root == null)
            return results;
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        int currentNumber = 1;
        int nextNumber = 0;
        while (queue.size() != 0) {
            TreeNode temp = queue.poll();
            result.add(temp.val);
            currentNumber--;
            if (temp.left != null) {
                queue.add(temp.left);
                nextNumber++;
            }
            if (temp.right != null) {
                queue.add(temp.right);
                nextNumber++;
            }
            if (currentNumber == 0) {
                results.add(result);
                currentNumber = nextNumber;
                nextNumber = 0;
                result = new ArrayList<Integer>();
            }
        }
        return results;
    }
    /*
     * 222.Count Complete Tree Nodes 
     * 11.20 By Mingyang
     * 最先各自计算最左边那一束到底的长度，再计算右边到底的长度，如果相等，ok，# of nodes = 2^h -1（计算满树很简单的！）
     * 若不相等，再继续迭代，这样做节省了很多时间
     * T(n)=logn+2T(n/2)，最后算出来就是n，其实可以这么理解，每个节点都只是访问了一次，所以还是n。注意树的高度是logn
     */
    public int countNodes(TreeNode root) {
        if (root == null)
            return 0;
        int leftHeight = 1, rightHeight = 1;
        // 计算左子树
        TreeNode temp = root.left;
        while (temp != null) {
            temp = temp.left;
            leftHeight++;
        }
        // 计算右子树
        temp = root.right;
        while (temp != null) {
            temp = temp.right;
            rightHeight++;
        }
        if (leftHeight == rightHeight)
            return (1 << leftHeight) - 1;
        // 也可以：(2<<(left-1))-1，这里h是left-1
        return countNodes(root.left) + countNodes(root.right) + 1;
    }
    /*
     * 226. Invert Binary Tree 
     * 11.19 By Mingyang
     */
     public TreeNode invertTree(TreeNode root){
            exchange(root);
            return root;
        }
        public void exchange(TreeNode root) {
           if(root==null)
             return;
           if(root.left==null&&root.right==null)
             return;
           TreeNode Temp=root.left;
           root.left=root.right;
           root.right=Temp;
           exchange(root.left);
           exchange(root.right);
        }
    /*
     * 230.Kth Smallest Element in a BST 
     * 11.19 By Mingyang
     */
    private static int number = 0;
    private static int county = 0;
    public int kthSmallest1(TreeNode root, int k) {
        county = k;
        dfs(root);
        return number;
    }
    public void dfs(TreeNode root) {
        if (root == null)
            return;
        dfs(root.left);
        county--;
        if (county == 0) {
            number = root.val;
            return;
        }
        dfs(root.right);
    }
    public int kthSmallest(TreeNode root, int k) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = root;
        int result = 0;
        // 不断的dfs，利用左中右的遍历方式进行遍历，然后依次取到最后的值
        while (!stack.isEmpty() || p != null) {
            if (p != null) {
                stack.push(p);
                p = p.left;
            } else {
                TreeNode t = stack.pop();
                k--;
                if (k == 0)
                    result = t.val;
                p = t.right;
            }
        }
        return result;
    }
    // 我的方法更快,利用左子树和右字数的个数的关系，如果k大于左子树的个数，那么就可以往右边走，这种方法复杂度T(n)=2T(n/2)+logn,也就是n
    //这里用了Binary Search的方法，根据count的大小来判断在哪个区间，然后去那个区间search，所以按理来说应该是logn，可是因为count用了n，所以总的时间就是n，哈哈！
    public int kthSmallest2(TreeNode root, int k) {
        int count = countNodes2(root.left);
        if (k <= count) {
            return kthSmallest2(root.left, k);
        } else if (k > count + 1) {
            return kthSmallest2(root.right, k-1-count); // 1 is counted as current node
        }
        return root.val;
    }
    public int countNodes2(TreeNode n) {
        if (n == null) return 0;

        return 1 + countNodes2(n.left) + countNodes2(n.right);
    }
    /*
     * 235. Lowest Common Ancestor of a Binary Search Tree 
     * 11.19 By Mingyang
     * 这题目更简单，下面两种方法都适用，现在给出更直接的方法
     */
    public TreeNode lowestCommonAncestor3(TreeNode root, TreeNode p, TreeNode q) {
        if (p.val == root.val || q.val == root.val)
            return root;
        if ((p.val < root.val && q.val > root.val)|| (p.val > root.val && q.val < root.val))
            return root;
        if (p.val < root.val) {
            return lowestCommonAncestor(root.left, p, q);
        } else {
            return lowestCommonAncestor(root.right, p, q);
        }
    }
    /*
     * 236. Lowest Common Ancestor of a Binary Tree 
     * 11.19 By Mingyang
     * 这道题目比BST更难，不过这道题目的答案可以通用的
     */
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // 发现目标节点则通过返回值标记该子树发现了某个目标结点
        if (root == null || root == p || root == q)
            return root;
        // 查看左子树中是否有目标结点，没有为null
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        // 查看右子树是否有目标节点，没有为null
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        // 都不为空，说明做右子树都有目标结点，则公共祖先就是本身
        if (left != null && right != null)
            return root;
        // 如果发现了目标节点，则继续向上标记为该目标节点
        return left == null ? right : left;
    }
    /*
     * 298 Binary Tree Longest Consecutive Sequence
     *  2016-3-27 by Mingyang
     *  The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections. 
     *  The longest consecutive path need to be from parent to child (cannot be the reverse).
     */
    private int result = Integer.MIN_VALUE;
    public int longestConsecutive(TreeNode root) {
        if(root == null)
            return 0;
        dfs(root, 0, root);
        return result;
    }
    private void dfs(TreeNode root, int cur, TreeNode pre) {
        if(root == null)
            return;
        if(root.val == pre.val+1)
            cur++;
        else
            cur = 1;
        result = Math.max(result, cur);
        dfs(root.left, cur, root);
        dfs(root.right, cur, root);
    }
    /*
     * 333.Largest BST Subtree
     * 2016-3-27 by Mingyang
     * Given a binary tree, find the largest subtree which is a Binary Search Tree (BST)
     * where largest means subtree with largest number of nodes in it.
     * why not traverse up the tree using a bottom-up approach? 
     * In other words, we verify the deeper nodes before we verify if the above nodes satisfy the BST requirements.
     * Using a bottom-up approach, we need to pass some information up the tree. 
     * Obviously, we need to pass minimum and maximum values of the subtree as we traverse up the tree, so the above subtrees could be verified for BST’s properties
     * 自底向上的方法，不过需要存lower，upper的值以及size的大小
     */
    class Result {  // (size, rangeLower, rangeUpper) -- size of current tree, range of current tree [rangeLower, rangeUpper]
        int size;
        int lower;
        int upper;
        Result(int size, int lower, int upper) {
            this.size = size;
            this.lower = lower;
            this.upper = upper;
        }
    }
    int max = 0;
    public int largestBSTSubtree(TreeNode root) {
        if (root == null) { return 0; }    
        traverse(root, null);
        return max;
    }
    private Result traverse(TreeNode root, TreeNode parent) {
        if (root == null) { return new Result(0, parent.val, parent.val); }
        Result left = traverse(root.left, root);
        Result right = traverse(root.right, root);
        if (left.size==-1 || right.size==-1 || root.val<left.upper || root.val>right.lower) {
            return new Result(-1, 0, 0);
        }
        int size = left.size + 1 + right.size;
        max = Math.max(size, max);
        return new Result(size, left.lower, right.upper);//更新新的BST的范围，新的Size
    }
}
/*
 * 297. Serialize and Deserialize Binary Tree 
 * 2016-3-27 by Mingyang
 */
class Codec {
     private static final String spliter = ",";
        private static final String NN = "X";
        // Encodes a tree to a single string.
        public String serialize(TreeNode root) {
            StringBuilder sb = new StringBuilder();
            buildString(root, sb);
            return sb.toString();
        }
        private void buildString(TreeNode node, StringBuilder sb) {
            if (node == null) {
                sb.append(NN).append(spliter);
            } else {
                sb.append(node.val).append(spliter);
                buildString(node.left, sb);
                buildString(node.right,sb);
            }
        }
        // Decodes your encoded data to tree.
        public TreeNode deserialize(String data) {
            Deque<String> nodes = new LinkedList<String>();
            nodes.addAll(Arrays.asList(data.split(spliter)));
            return buildTree(nodes);
        }
        private TreeNode buildTree(Deque<String> nodes) {
            String val = nodes.remove();
            if (val.equals(NN)) return null;
            else {
                TreeNode node = new TreeNode(Integer.valueOf(val));
                node.left = buildTree(nodes);
                node.right = buildTree(nodes);
                return node;
            }
        }
}