思路:递归解决,在返回root前保证该点的两个孩子已经互换了.注意可能给一个Null. C++ /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeN…

Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia: This problem was inspired by this original tweet by Max Howell: Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tre…

翻转一棵二叉树. 示例: 输入: 4 / \ 2 7 / \ / \ 1 3 6 9 输出: 4 / \ 7 2 / \ / \ 9 6 3 1  思路 如果根节点存在,就交换两个子树的根节点,用递归,从下至上.. 解一: auto tmp = invertTree(root->left); 将左子树整体 当作一个节点 交换. 最后返回根节点. /** * Definition for a binary tree node. * struct TreeNode { * int val; * Tr…

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* invertTree(TreeNode* root) { if (root == NU…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. Example Given 4 points: (1,2), (3,6), (0,0), (1,3). The maximum number is 3. LeeCode上的原题,可参见我之前的博客Invert Binary Tree 翻转二叉树. 解法一: // Recursion class So…

LeetCode-- Invert Binary Tree Question invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia: This problem was inspired by this original tweet by Max Howell: Google: 90% of our engineers use the software you wrote (Homeb…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 迭代 日期 题目地址: https://leetcode.com/problems/invert-binary-tree/ 题目描述 Invert a binary tree. Example: Input: 4 / \ 2 7 / \ / \ 1 3 6 9 Output: 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia:…

大牛没有能做出来的题,我们要好好做一做 Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia: This problem was inspired by this original tweet by Max Howell: Google: 90% of our engineers use the software you wrote (Homebrew), but you can't…

Print a binary tree in an m*n 2D string array following these rules: The row number m should be equal to the height of the given binary tree. The column number n should always be an odd number. The root node's value (in string format) should be put i…

［抄题］: Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 [暴力解法]: 时间分析: 空间分析: ［奇葩输出条件］: ［奇葩corner case］: ［思维问题］: 以为要分为r.r = l.l, r.l = l.r来讨论,但是其实这样只能最后判断相等.翻转是每一步都要进行的动作,应该尽早开始 ［一句话思路］: ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 给一个二叉树,返回所有根到叶节点的路径. Java: /** * Definition for a binary tree node…

找到所有根到叶子的路径 深度优先搜索(DFS), 即二叉树的先序遍历. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<string> v…

题目: 翻转二叉树 翻转一棵二叉树 样例 1 1 / \ / \ 2 3 => 3 2 / \ 4 4 挑战 递归固然可行,能否写个非递归的? 解题: 递归比较简单,非递归待补充 Java程序: /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val;…

Given 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"] Explanation: All root-to-leaf paths are: 1->2->5, 1->3 题目 给定一棵二叉树,…

Description: Invert a binary tree. 4    /    \  2      7 /  \    /   \1   3   6   9 to 4 / \ 7 2 / \ / \ 9 6 3 1递归invert就好了. /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(i…

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public TreeNode invertTree(TreeNode root) { if (root == null) return null;…

# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def invertTree(self, root): """ :type root: TreeNode :rtype: TreeNode &quo…

Given an integer array with no duplicates. A maximum tree building on this array is defined as follow: The root is the maximum number in the array. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number…

struct TreeNode* invertTree(struct TreeNode* root) { if ( NULL == root ) { return NULL; } if ( NULL == root->left && NULL == root->right ) { //叶子节点 return root; } //交换左右子节点 struct TreeNode * pTreeNodeTmp = root->left; root->left = root…

1 题目 Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 接口: public TreeNode invertTree(TreeNode root) 2 思路 反转一颗二叉树. 可以用递归和非递归两种方法来解. 递归的方法,写法非常简洁,五行代码搞定,交换当前左右节点,并直接调用递归即可. 非递归的方法,参考树的层序遍历,借助Queue来辅助,先把根节点排入队列中,然后从队中取出来,交换其左…

leetcode 226. Invert Binary Tree 倒置二叉树 思路:分别倒置左边和右边的结点,然后把根结点的左右指针分别指向右左倒置后返回的根结点. # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): d…

226. Invert Binary Tree Total Accepted: 57653 Total Submissions: 136144 Difficulty: Easy Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia: This problem was inspired by this original tweet by Max Howell: Google: 90%…

翻译 给定一个二叉树,返回从下往上遍历经过的每一个节点的值. 从左往右,从叶子到节点. 比如: 给定的二叉树是 {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 返回它从下往上的遍历结果: [ [15,7], [9,20], [3] ] 原文 Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, lev…

题目: Invert a binary tree. 翻转二叉树. 递归,每次对节点的左右节点调用invertTree函数,直到叶节点. python中也没有swap函数,当然你可以写一个,不过python中可以通过:a, b = b, a交换两个变量的值 class Solution(object): def invertTree(self, root): if root == None: return root root.left, root.right = self.invertTree(r…

258. Add Digits Digit root 数根问题 /** * @param {number} num * @return {number} */ var addDigits = function(num) { var b = (num-1) % 9 + 1 ; return b; }; //之所以num要-1再+1;是因为特殊情况下:当num是9的倍数时,0+9的数字根和0的数字根不同. 性质说明 1.任何数加9的数字根还是它本身.(特殊情况num=0)        小学学加法的…

翻译 给定一个二叉树.返回其兴许遍历的节点的值. 比如: 给定二叉树为 {1. #, 2, 3} 1 \ 2 / 3 返回 [3, 2, 1] 备注:用递归是微不足道的,你能够用迭代来完毕它吗? 原文 Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: R…

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 or…

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 树的遍历,最常见的有先序遍历,中序遍历,后序遍历和层序遍历,它们用递归实现起来都非常的简…

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. Example: Input: [1,2,3,null,5,null,4] Output: [1, 3, 4] Explanation: 1 <--- / \ 2 3 <--- \ \ 5 4 <---  …

Given a binary tree, find the length of the longest consecutive sequence path. The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The longest consecutive path need to be from p…