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,...
原题网址(中):https://leetcode-cn.com/problems/binary-tree-pruning/ 原题网址(英):https://leetcode.com/problems/binary-tree-pruning/ 题目: We are given the head node root of a binary tree, where additionally every node’s value is either a 0 or a 1. Return the same tree where every...
pruneTree(root.right) # 如果当前结点的值为 1 或者左/右子树存在, # 则当前子树必定含有值为 1 的子结点,直接返回 root 即可 if root.val == 1 or root.left or root.right: return root # 此时当前子树中的所有子结点都是 0 ,需要修剪掉 return None 代码(Go) /** * Definition for a binary ...
* public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class ...
LeetCode 102. 二叉树的层序遍历 Binary Tree Level Order Traversal (广度优先搜索(BFS))...,102.二叉树的层序遍历给你一个二叉树,请你返回其按层序遍历得到的节点值。(即逐层地,从左到右访问所有节
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)。
https://leetcode.com/problems/binary-tree-inorder-traversal/ 2. 分析 2.1 迭代法 class Solution { public: vector<int> inorderTraversal(TreeNode* root) { vector<int> ans; stack<TreeNode*> todo;//定义一个栈,先入后出 ...
链接:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/ 难度:Medium 题目:105. Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a tree, construct the binary tree. ...
通过率 击败用户 0% 击败用户 0% 击败用户 0% 0尝试中 0次提交 0尝试中 0尝试中 0尝试中 简单 0/52 中等 0/149 困难 0/23 中等 96. 不同的二叉搜索树 中等 98. 验证二叉搜索树 中等 99. 恢复二叉搜索树 中等 100. 相同的树 简单 101. 对称二叉树 ...
classSolution{publicbooleanisCompleteTree(TreeNoderoot){Queue<TreeNode>queue=newLinkedList<>();TreeNodeprev=root;queue.add(root);while(!queue.isEmpty()){TreeNodenode=queue.remove();if(prev==null&&node!=null)returnfalse;if(node!=null){queue.add(node.left);queue.add(node.right);}prev=...