Leetcode: Binary Tree Level Order Transversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},
    3
   / 
  9  20
    /  
   15   7
return its bottom-up level order traversal as:
[
  [15,7],
  [9,20],
  [3]
]

第二遍方法：

这道题在groupon面经里面有，有一个follow up 是能不能右对齐输出。那就在29行记录每一行的最大size，然后在输出的时候根据最大size补齐空格

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (root == null) return res;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> item = new ArrayList<Integer>();
            int size = queue.size();
            for (int i=0; i<size; i++) {
                TreeNode cur = queue.poll();
                item.add(cur.val);
                if (cur.left != null) {
                    queue.add(cur.left);
                }
                if (cur.right != null) {
                    queue.add(cur.right);
                }
            }
            res.add(0, new ArrayList<Integer>(item));
        }
        return res;
    }
}

在Binary Tree Level Order Transversal的基础上难度：20，只需要对最后结果做一个倒序就好。格式是Collections.reverse(List<?> list)

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
        ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>> ();
        if (root == null) return lists;
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        int ParentNumInQ = 1;
        int ChildNumInQ = 0;
        ArrayList<Integer> list = new ArrayList<Integer>();
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            list.add(cur.val);
            ParentNumInQ--;
            if (cur.left != null) {
                queue.add(cur.left);
                ChildNumInQ++;
            }
            if (cur.right != null) {
                queue.add(cur.right);
                ChildNumInQ++;
            }
            if (ParentNumInQ == 0) {
                ParentNumInQ = ChildNumInQ;
                ChildNumInQ = 0;
                lists.add(list);
                list = new ArrayList<Integer>();
            }
        }
        Collections.reverse(lists);
        return lists;
    }
}

 注意38行的写法，Collections.reverse()跟Arrays.sort()函数一样，都是void返回型，然后改变作用在argument上

还有if (root == null) return null;

ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();这样会出错：

Input:{}Output:nullExpected:[]

DFS 做法：

public class Solution {
        public List<List<Integer>> levelOrderBottom(TreeNode root) {
            List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
            levelMaker(wrapList, root, 0);
            return wrapList;
        }
        
        public void levelMaker(List<List<Integer>> list, TreeNode root, int level) {
            if(root == null) return;
            if(level >= list.size()) {
                list.add(0, new LinkedList<Integer>());
            }
            levelMaker(list, root.left, level+1);
            levelMaker(list, root.right, level+1);
            list.get(list.size()-level-1).add(root.val);
        }
    }