#include <stdio.h> #include <memory.h> #include <math.h> #include <string> #include <vector> #include <set> #include <stack> #include <queue> #include <algorithm> #include <map> #define I scanf #…

背景描述 Homebrew 是 OS X 平台上的包管理工具.用其官网的话说就是: the missing package manager for OS X | OS X 平台遗失的包管理器. 相信在用 Mac 的程序员,差不多都知道 Homebrew. Homebrew 的开发者是 Max Howell.今天 Max 在推特发帖: Google: 90% of our engineers use the software you wrote (Homebrew), but you can't…

1.题目描述 经典的反转二叉树,就是将二叉树中每个节点的左.右儿子交换. 2.题目分析 3.代码 TreeNode* invertTree(TreeNode* root) { if(root == NULL ) return NULL; TreeNode* temp; temp = root->left; root->left = invertTree(root->right); root->right = invertTree(temp); return root; }…

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 tree…

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 tree…

// 二叉树节点定义 public class BinaryTreefanzhuan { class TreeNode{ int value; TreeNode left; TreeNode right; } // 递归 public static TreeNode invertNode(TreeNode root){ if (root == null) return null;  TreeNode temp = root.left; root.left = invertNode(root.ri…

［抄题］: Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes.…

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…

题目描述: 翻转一棵二叉树. 解题思路: 1.对于二叉树,立马递归 2.先处理 根节点,不需改动 3.处根的左子树和右子树需要交换位置 4.递归处理左子树和右子树.步骤见1-3步 Java代码实现: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */…

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…