LeetCode:Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal II

Total Accepted: 84098 Total Submissions: 244151 Difficulty: Easy

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,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

[  [15,7],  [9,20],  [3]]

Subscribe to see which companies asked this question

Hide Tags

 Tree Breadth-first Search

Hide Similar Problems

 (E) Binary Tree Level Order Traversal

思路：

承接上题【Binary Tree Level Order Traversal】，将结果的插入方式改为“头插”即可。

java code:

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