link:[https://leetcode.com/explore/learn/card/data-structure-tree/134/traverse-a-tree/928/] 递归解法: #Definition for a binary tree node.#class TreeNode(object):#def __init__(self, x):#self.val = x#self.left = None#self.right = NoneclassSolution(object):defsolve(self,root):ifroo...
Left != nil || root.Right != nil { return root } // 此时当前子树中的所有子结点都是 0 ,需要修剪掉 return nil } 题目链接: Binary Tree Pruning: leetcode.com/problems/b 二叉树剪枝: leetcode.cn/problems/bi LeetCode 日更第 235 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
https://leetcode.com/problems/binary-tree-postorder-traversal/ Given a binary tree, return thepostordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[3,2,1]. Note:Recursive solution is trivial, could you do it iteratively? 后续遍历二叉树,...
Python 代码 栈 + 循环实现 # Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def binaryTreePaths(self, root: TreeNode) -> List[str]: res = ...
Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. Note: A leaf is a node with no children. Example 1: Input: root = [3,9,20,null,null,15,7] Output: 2 Example 2: ...
Can you solve this real interview question? Lowest Common Ancestor of a Binary Tree - Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia [https://en.wikipedia.org/wi
中等 96. 不同的二叉搜索树 71.5% 中等 98. 验证二叉搜索树 39.3% 中等 99. 恢复二叉搜索树 61.3% 中等 100. 相同的树 63.2% 简单 101. 对称二叉树 62.3% 简单 102. 二叉树的层序遍历 69.7% 中等 103. 二叉树的锯齿形层序遍历 60.2% 中等 104. 二叉树的最大深度 78.6% 简单 105. 从前序与中序遍历...
968. 监控二叉树 - 给定一个二叉树,我们在树的节点上安装摄像头。 节点上的每个摄影头都可以监视其父对象、自身及其直接子对象。 计算监控树的所有节点所需的最小摄像头数量。 示例 1: [https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/29/bst_c
图片来源:https://leetcode.com/articles/binary-tree-right-side-view/ 对于时间复杂度来说,基本上都是O(n),因为要访问所有的点。 对于空间复杂度来说,BFS取决于扫描过程中每层的node数,就是树的宽度,而DFS取决于扫描过程中树的深度。最坏情况两个都是O(n)。
http://bangbingsyb.blogspot.com/2014/11/leetcode-balanced-binary-tree.html 我想说明的是,什么 depth of tree and height of tree depth of node n: length of path from n to root. 所以说, depth of subtree 指的应该就是 这棵subtree最大的深度,以该subtree结点为root。