否则,我们应访访问当前节点的右孩子(入栈)。 // C++ CODE: classSolution {public: vector<int> postorderTraversal(TreeNode*root) { stack<TreeNode*>s; vector<int>result;while(root){ s.push(root); root= root->left; }while(!s.empty(
#Definition for a binary tree node.#class TreeNode(object):#def __init__(self, x):#self.val = x#self.left = None#self.right = None"""利用堆栈先进后出的特点"""classSolution(object):defpreorderTraversal(self, root):""":type root: TreeNode :rtype: List[int]"""result=[] slack=...
[LeetCode] 二叉树的前序,后序,中序,层序的递归和迭代 powcai 「leetcode」236. 二叉树的最近公共祖先【递归】详解! 236. 二叉树的最近公共祖先链接: https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。 百度百科中最近公...
LeetCode 110. Balanced Binary Tree Description Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the depth of the two subtrees of every node never differ by more than 1. Example 1: Given the ...
Given a binary tree, return thebottom-up level ordertraversal 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 ...
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
Can you solve this real interview question? Binary Tree Preorder Traversal - Given the root of a binary tree, return the preorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,2,3] Explanation: [https://assets.l
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)。
Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 思路: 翻转二叉树,刚开始陷入了一个思维怪圈,认为先正向保存值至一个列表,然后再按照逆向还原这个链表,这种思路太复杂,如果是对称二叉树还好,非对称的就不好做。最后看了下别人的解题答案,很简单。先逆...
Leetcode每日一题:110.balanced-binary-tree(平衡二叉树),思路:依旧是递归,这里暂时只想到了比较笨的方法,就是对从上至下对当前结点的左右孩子,遍历得出它们高度,然后做差,判断是否满足平衡二叉树的条件;想过动态规划和从下至上的思想,但技术没到位,没能实现出来