}publicstaticvoidmain(String[] args){BinaryTreeInOrderTraversalbinaryTreeInOrderTraversal=newBinaryTreeInOrderTraversal();char[] arr1 =newchar[]{'1','#','2','3'};char[] arr2 =newchar[]{'1','2','3','#','#','4','#','#','5'}; System.out.println(Arrays.toString(binaryTreeI...
package leetcode; import java.util.*; //方法一 数据结构:树的中序遍历:1、递归 2、非递归 //始终操作的是根节点 ->( root=root.left),把右边节点移到根节点上进行; public class BinaryTreeInorderTraversal { private class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) {...
94. Binary Tree Inorder TraversalEasy Topics Companies Given the root of a binary tree, return the inorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,3,2] Explanation: Example 2: Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] ...
102. 二叉树的层序遍历 - 给你二叉树的根节点 root ,返回其节点值的 层序遍历 。 (即逐层地,从左到右访问所有节点)。 示例 1: [https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg] 输入:root = [3,9,20,null,null,15,7] 输出:[[3],[9,20],[15,7]]
名字叫做, morrois traversal, 自己写了下: My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>inorderTraversal(TreeNoderoot){List...
这道题和LeetCode笔记:107. Binary Tree Level Order Traversal II是姊妹题,解题思路都是一样的,只是结果要求的顺序是反的,同样有两种方法,也就是经常说到的DFS深度优先遍历和BFS广度优先遍历。 BFS: 广度优先遍历就是一层层地攻略过去,把每一层的所有节点都记录下来再走向下一层。因为每层会有多个节点,不是简...
105. 从前序与中序遍历序列构造二叉树 - 给定两个整数数组 preorder 和 inorder ,其中 preorder 是二叉树的先序遍历, inorder 是同一棵树的中序遍历,请构造二叉树并返回其根节点。 示例 1: [https://assets.leetcode.com/uploads/2021/02/19/tree.jpg] 输入: preo
Can you solve this real interview question? Construct Binary Tree from Inorder and Postorder Traversal - Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of th
The function returns the root of the constructed binary tree. 4. Time & Space Complexity Analysis: 4.1 Time Complexity: 4.1.1 Lists as parameters In each recursive call, the index() function is used to find the index of the root value in the inorder traversal list. This function has a ...
Leetcode 94: Binary Tree Inorder Traversal-中序遍历 自动驾驶搬砖师 一步一个台阶,一天一个脚印 一、题目描述 给定一个二叉树,返回它的中序遍历。 二、解题思路 2.1 递归法 时间复杂度 O(n) 空间复杂度 O(n) 2.2 迭代法 时间复杂度 O(n) 空间复杂度 O(n)(1) 同理创建一个Stack,然后按左/中/右...