本题也属于层次遍历的变形,不同之处在于其遍历的方法是交替进行的,形成一个ZigZag的曲线形式,如下: 代码如下: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 1 struct TreeNode { 2 int val; 3 TreeNode* left; 4 TreeNode* right; 5 TreeNode(int x): val(x), left(N
classSolution {public: vector<vector<int> > zigzagLevelOrder(TreeNode *root) { vector<vector<int>>ans;if(root ==NULL) {returnans; }intlevel =0;//记录当前层号vector<TreeNode *>curlevel; curlevel.push_back(root);while(!curlevel.empty()) { vector<int>partans; vector<TreeNode *>nextl...
Given a binary tree,returnthe zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).For example: Given binary tree [3,9,20,null,null,15,7],3 /\9 20 /\15 7returnits zigzag level order traversal as...
236. 二叉树的最近公共祖先链接: https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。 百度百科中最近公… 代码随想录发表于数据结构与...打开知乎App 在「我的页」右上角打开扫一扫 其他扫码方式:微信 下载知乎App 开通机构号...
【LeetCode】二叉树的层次之字形遍历:Binary Tree Zigzag Level Order Traversal,代码先锋网,一个为软件开发程序员提供代码片段和技术文章聚合的网站。
[Leetcode][python]Binary Tree Zigzag Level Order Traversal,题目大意按之字形遍历二叉树(一正一反)解题思路来自:链接解题思路:这道题和上一题层序遍历那道题差不多,区别只是在于奇数层的节点要翻转过来存入数组。代码:代码BFSclassSolution(object):defzigzagLeve
LeetCode 103. Binary Tree Zigzag Level Order Traversal 原题链接在这里:https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ 题目: Given a binary tree, return thezigzag level ordertraversal of its nodes' values. (ie, from left to right, then right to left for the next ...
Can you solve this real interview question? Longest ZigZag Path in a Binary Tree - You are given the root of a binary tree. A ZigZag path for a binary tree is defined as follow: * Choose any node in the binary tree and a direction (right or left). *
题源:LeetCode 【Binary Tree 系列最终章】 这篇文章汇总了数据结构二叉树 (Binary Tree) 相关问题的多种解法。针对简单题目,讨论的重点倾向于对Python编程知识的活学活用,和思路的发散与实现。 文中第三题,用中序遍历和前序遍历验证二叉搜索树是本篇的重点。
这道题目,我怎么就没有想到,拿队列来做呢。然后采用zigzag的方式遍历。 http://www.programcreek.com/2014/04/leetcode-binary-tree-right-side-view-java/ 看了这篇博客就懂了。 ** 总结: queue来遍历tree ** Anyway, Good luck, Richardo! My code: ...