二叉树

一、二叉树与递归

        二叉树具有天然的递归性。

        关于递归函数的写法要明确的几个点：1.递归出去的条件（开头一个return （是递归出去条件）  结尾处return（是返回执行的结果）） 2.明确函数的实际意义（以便于后面可以直接按照含义来递归执行） 3.递归过程要执行的逻辑

1.以下简单的递归都是 1).递归出去的条件 +  2).明确函数的实际意义 就可以直接构建出递归函数来解决问题

104. Maximum Depth of Binary Tree

Given binary tree [3,9,20,null,null,15,7],

    3
   / 
  9  20
    /  
   15   7

return its depth = 3.

代码如下：

class Solution {
   public int maxDepth(TreeNode root) {
        if(root == null) {
            return 0;
        }
        else{
            return 1+ Math.max(maxDepth(root.left), maxDepth(root.right));
        }
    }
}

111. Minimum Depth of Binary Tree

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3
   / 
  9  20
    /  
   15   7

return its minimum depth = 2.

代码如下：

//为什么采用分治的思想而不是求数最大深度思想，因为可能为单链表（链表是特殊树）没法处理
class Solution {
     public int minDepth(TreeNode root) {
        if(root == null) return 0;
        int left = minDepth(root.left);
        int right = minDepth(root.right);
        return (left == 0 || right == 0) ? left + right + 1//为什么可以写成left + right + 1
            : Math.min(left,right) + 1;         //因为left 跟right 必有一个为0，所以..
       
    }
}

226. Invert Binary Tree

Invert a binary tree.

Example:

Input:

     4
   /   
  2     7
 /    / 
1   3 6   9

Output:

     4
   /   
  7     2
 /    / 
9   6 3   1


代码如下：

public class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null) return null; //递归出口
        TreeNode node = root.left;
        root.left = invertTree(root.right);
        root.right = invertTree(node);
        return root;  //执行结果
    }
}

//还有另一种递归思想 root的左子树先反转 root的右子树也先反转， 后面再交换root的左右子树

100. Same Tree

Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Example 1:

Input:     1         1
          /        / 
         2   3     2   3

        [1,2,3],   [1,2,3]

Output: true

代码如下：

class Solution {
   public boolean isSameTree(TreeNode p, TreeNode q) {
    if(p == null && q == null) return true;
    if(p == null || q == null) return false;//递归出去条件有两
    if(p.val == q.val)
        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    return false;
  }
}

101. Symmetric Tree

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / 
  2   2
 /  / 
3  4 4  3


代码如下：

class Solution {
  public boolean isSymmetric(TreeNode root) {
    if(root==null) return true;
    return isMirror(root.left,root.right);
}
public boolean isMirror(TreeNode p, TreeNode q) {
    if(p==null && q==null) return true;
    if(p==null || q==null) return false;
    return (p.val==q.val) && isMirror(p.left,q.right) && isMirror(p.right,q.left);
   }
}

222. Count Complete Tree Nodes

Given a complete binary tree, count the number of nodes.

Note:

Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2hnodes inclusive at the last level h.

Example:

Input: 
    1
   / 
  2   3
 /   /
4  5 6

Output: 6
分析：

1. 假如左子树高度等于右子树高度，则右子树为完全二叉树，左子树为满二叉树。
2. 假如高度不等，则左字数为完全二叉树，右子树为满二叉树。
3. 求高度的时候只往左子树来找。

代码如下：

class Solution {  
    public int countNodes(TreeNode root) {
        int h = height(root);
        return h < 0 ? 0 :
               height(root.right) == h-1 ? (1 << h) + countNodes(root.right)
                                         : (1 << h-1) + countNodes(root.left);
    }
    
   int height(TreeNode root) {
        return root == null ? -1 : 1 + height(root.left);
    } 
}

//或者直接对任意二叉树用暴力解法
class Solution {
        public int countNodes(TreeNode* root) {
               return root ? (1 + countNodes(root->left) + countNodes(root->right)) : 0;
       }
}

110. Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / 
  9  20
    /  
   15   7

Return true.

代码如下：

public class Solution {
   private boolean result = true;

   public boolean isBalanced(TreeNode root) {
      maxDepth(root);
      return result;
}

   public int maxDepth(TreeNode root) {
      if (root == null) return 0;
      int l = maxDepth(root.left);
      int r = maxDepth(root.right);
      if (Math.abs(l - r) > 1)
          result = false;
      return 1 + Math.max(l, r);
  }
}

2. 递归终止条件存在陷阱的递归

112. Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / 
    4   8
   /   / 
  11  13  4
 /        
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

特殊情况 二叉树如下 sum = 5

      5
       
        8
       /  
      13   4
        
   

代码如下：

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
    
        if(root.left == null && root.right == null && sum - root.val == 0) return true; //判断为叶子节点
    
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

404. Sum of Left Leaves（跟上一题一样，判断是否为叶子节点）

Find the sum of all left leaves in a given binary tree.

Example:

    3
   / 
  9  20
    /  
   15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

代码如下：

class Solution {
public int sumOfLeftLeaves(TreeNode root) {
    if(root == null) return 0;
    int ans = 0;
    if(root.left != null) {
        if(root.left.left == null && root.left.right == null) ans += root.left.val;//判断是否为叶子节点
        else ans += sumOfLeftLeaves(root.left);
    }
    ans += sumOfLeftLeaves(root.right);
    
    return ans;
   }
}

3.执行逻辑部分比较复杂的二叉树的递归

257. Binary Tree Paths

iven a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   
2     3
 
  5

Output: ["1->2->5", "1->3"]

class Solution {
   public List<String> binaryTreePaths(TreeNode root) {
      List<String> answer = new ArrayList<String>();
      if (root != null) searchBT(root, "", answer);
      return answer;
}
  private void searchBT(TreeNode root, String path, List<String> answer) {
      if (root.left == null && root.right == null) answer.add(path + root.val);
      if (root.left != null) searchBT(root.left, path + root.val + "->", answer);
      if (root.right != null) searchBT(root.right, path + root.val + "->", answer);
  }
}

113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / 
    4   8
   /   / 
  11  13  4
 /      / 
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

代码如下：

class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum){
        List<List<Integer>> result  = new LinkedList<List<Integer>>();
        List<Integer> currentResult  = new LinkedList<Integer>();
        pathSum(root,sum,currentResult,result);
        return result;
}

    public void pathSum(TreeNode root, int sum, List<Integer> currentResult,
        List<List<Integer>> result) {

        if (root == null)
            return;
        currentResult.add(new Integer(root.val));
        if (root.left == null && root.right == null && sum == root.val) {
            result.add(new LinkedList(currentResult));
            currentResult.remove(currentResult.size() - 1);//don't forget to remove the last integer
            return;
        } else {
            pathSum(root.left, sum - root.val, currentResult, result);
            pathSum(root.right, sum - root.val, currentResult, result);
        }
        currentResult.remove(currentResult.size() - 1);
    }
}

129. Sum Root to Leaf Numbers

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

Note: A leaf is a node with no children.

Input: [4,9,0,5,1]
    4
   / 
  9   0
 / 
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

代码如下：

class Solution {
    public int sumNumbers(TreeNode root) {
    return sum(root, 0);
}

public int sum(TreeNode n, int s){
    if (n == null) return 0;
    if (n.right == null && n.left == null) return s*10 + n.val;
    return sum(n.left, s*10 + n.val) + sum(n.right, s*10 + n.val);
   }
}

其实以上递归出去的条件基本是 root = null,      n.right == null && n.left == null (叶子节点)     n.right == null || n.left == null 

437. Path Sum III

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  
    5   -3
   /     
  3   2   11
 /    
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

代码如下：

public class Solution {
    public int pathSum(TreeNode root, int sum) {
        if (root == null) return 0;
        return pathSumFrom(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }
    
    private int pathSumFrom(TreeNode node, int sum) {
        if (node == null) return 0;
        return (node.val == sum ? 1 : 0) 
            + pathSumFrom(node.left, sum - node.val) + pathSumFrom(node.right, sum - node.val);
    }
}

4.二分搜索树

235. Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

代码如下：

class Solution {
   public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
       while((root.val - p.val)*(root.val -q.val) > 0){
           root = root.val > p.val ? root.left : root.right;
       }
       return root;
   
   }
}

//或者
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root.val > p.val && root.val > q.val){
            return lowestCommonAncestor(root.left, p, q);
        }else if(root.val < p.val && root.val < q.val){
            return lowestCommonAncestor(root.right, p, q);
        }else{
            return root;
        }
    }
}

98. Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

Example 1:

    2
   / 
  1   3

Input: [2,1,3]
Output: true
验证该树是否为二分搜索树

 代码如下：

//可以中序遍历出来递增则是.....

//核心思想是左子树的最大值小于跟节点，右子树最小值大于跟节点
public class Solution {
    public boolean isValidBST(TreeNode root) {
        return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    
    public boolean isValidBST(TreeNode root, long minVal, long maxVal) {
        if (root == null) return true;
        if (root.val >= maxVal || root.val <= minVal) return false;
        return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
    }                //左子树的最大值就是跟节点 所以maxVal = root.val，
}

108. Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / 
   -3   9
   /   /
 -10  5

分析：有序数组转为二叉树其实就是对二叉树进行中序遍历的逆过程
代码如下：

class Solution {
   public TreeNode sortedArrayToBST(int[] num) {
    if (num.length == 0) {
        return null;
    }
    TreeNode head = helper(num, 0, num.length - 1);
    return head;
}

public TreeNode helper(int[] num, int low, int high) {
    if (low > high) { // Done
        return null;
    }
    int mid = (low + high)/2;
    TreeNode node = new TreeNode(num[mid]);
    node.left = helper(num, low, mid-1);
    node.right = helper(num, mid+1 ,high);
    return node;
    
}
}

230. Kth Smallest Element in a BST

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note: 
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Example 1:

Input: root = [3,1,4,null,2], k = 1
   3
  / 
 1   4
  
   2
Output: 1

代码如下：

class Solution {
   public int kthSmallest(TreeNode root, int k) {
      Stack<TreeNode> st = new Stack<>();
      
      while (root != null) {
          st.push(root);
          root = root.left;
      }
          
      while (k != 0) {
          TreeNode n = st.pop();
          k--;
          if (k == 0) return n.val;
          TreeNode right = n.right;
          while (right != null) {
              st.push(right);
              right = right.left;
          }
      }
      
      return -1; // never hit if k is valid
   }
}

236. Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

分析：任意一颗二叉树的公共祖先
代码如下：

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null || root == p || root ==q) return root;
        TreeNode left = lowestCommonAncestor(root.left,p,q); //查找pq是否在左子树
        TreeNode right = lowestCommonAncestor(root.right,p,q);//
        if(left == null){
            return right;
        }
        if(right == null){
            return left;
        }
       return root;
    }
}