用栈模拟二叉树。 Python3代码 # Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = NoneclassSolution:definvertTree(self, root: TreeNode) -> TreeNode:# solution two: 栈ifnotroot:returnNone# 叶子节点,直接返回...
Python解Leetcode: 226. Invert Binary Tree 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...
226. Invert Binary Tree aliblielite 来自专栏 · leetcode_python_easy # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # recursively,树的问题大都很适合用递归,因为每一个节点都...
* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicTreeNodeinvertTree(TreeNode root){if(root==null){// 节点为空,不处理returnnull;}else{// 节点不为空Tree...
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off. 参见这:这段代码体现的思维,就是旧树到新树的映射——对一颗二叉树而言,它的镜像树就是左右节点递归镜像的树。这...
226. Invert Binary Tree 题目大意: 翻转二叉树 思路:递归一下,递归比迭代好写 Python代码: # 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):...
Tree - 144. Binary Tree Preorder Traversal binarylistnulltraversaltree Given a binary tree, return the preorder traversal of its nodes' values. ppxai 2020/09/23 3040 leetcode 226 Invert Binary Tree 翻转二叉树 编程算法python Trivia: This problem was inspired by this original tweet by Max Howe...
Python Solution: # Definitionfora binary tree node. #classTreeNode(object): # def __init__(self, x): # self.val=x # self.left=None # self.right=NoneclassSolution(object): def invertTree(self, root):""":type root: TreeNode
利用层序遍历 , 也就是队列 伪代码: make a queue push root into queue if queue is not empty node= queue pop if has left if left has child enqueue left if has right...
invertTree(root.right); }returnroot; } } Python: # Runtime: 52 ms# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = NoneclassSolution:# @param {TreeNode} root# @return {TreeNode}definvertTree(self, ...