# 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…

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…

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…

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…

原题地址:http://oj.leetcode.com/problems/balanced-binary-tree/ 题意:判断一颗二叉树是否是平衡二叉树. 解题思路:在这道题里,平衡二叉树的定义是二叉树的任意节点的两颗子树之间的高度差小于等于1.这实际上是AVL树的定义.首先要写一个计算二叉树高度的函数,二叉树的高度定义为:树为空时,高度为0.然后递归求解:树的高度 = max(左子树高度,右子树高度)+1(根节点要算上).高度计算函数实现后,递归求解每个节点的左右子树的高度差,如果有大于1的…

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;…

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…

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…