Leetcode: Binary Tree Level Order Transversal II Given a binary tree,returnthe 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 7returnits bottom-up...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<vector<int>> levelOrderBottom(TreeNode*root) { levelOrder(root,0); reverse(res.begin(), res.end());returnres; }private:voidlevelOrder(TreeNode* root,intcurrLevel) {if(root == NULL)re...
Given a binary tree, return thebottom-up level ordertraversal 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 1. 2. 3. 4. 5. return its bottom-up level orde...
这道题目跟上道题目很相似 Leetcode: Binary Tree Level Order Traversal ,唯一不同的就是返回结果是从子叶节点到根节点,所以我们只需要将结果翻转下就好了! 参考代码: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : ...
Can you solve this real interview question? Binary Tree Level Order Traversal - Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level). Example 1: [https://assets.leetcode.c
题目107. Binary Tree Level Order Traversal 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 exam...
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its level order traversal as: ...
102. 二叉树的层序遍历 - 给你二叉树的根节点 root ,返回其节点值的 层序遍历 。 (即逐层地,从左到右访问所有节点)。 示例 1: [https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg] 输入:root = [3,9,20,null,null,15,7] 输出:[[3],[9,20],[15,7]]
BFS with level-order traversal using a queue, for each node in a level, we swap its left child and right child; then, adding children to queue to be processed in next level. Solution The code is written in Java. Using DFS publicTreeNodeinvertTree(TreeNode root){if(root==null)returnroo...
1. 每进入一层用一个变量(这里是levelNum)记录当前从根节点到这一层节点的节点数 2. 当到达叶子节点的时候,就记录当前最小深度overall 2. 当所有的叶子节点都到达一遍之后,就得到了最终最小深度overall。 class Solution { public: int minDepth(TreeNode *root) { ...