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…

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…

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…

［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…

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;…

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…

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…

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…

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…

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…

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…

/** * 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 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…

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…

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…

Invert a binary tree 翻转一棵二叉树 假设有如下一棵二叉树: 4  / \   2    7  / \   / \ 1  3 6  9翻转后: 4     /    \    7     2   / \    / \  9  6  3  1 这里采用递归的方法来处理.遍历结点,将每个结点的两个子结点交换位置即可. 从左子树开始,层层深入,由底向上处理结点的左右子结点:然后再处理右子树 全部代码如下: public class InvertBinaryTree { public…

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%…

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…

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…

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 <---  …