TreeNode root =newTreeNode(4); root.left =newTreeNode(2); root.left.left =newTreeNode(1); root.left.right =newTreeNode(3); root.right =newTreeNode(6); root.right.left =newTreeNode(5); root.right.right =newTreeNode(7); root.right.right.right =newTreeNode(8); System.out.print...
请一键三连, 非常感谢LeetCode 力扣题解94. 二叉树的中序遍历94. Binary Tree Inorder Traversal帮你深度理解 二叉树 数据结构 递归算法 Morris遍历 迭代, 视频播放量 116、弹幕量 0、点赞数 0、投硬币枚数 0、收藏人数 0、转发人数 0, 视频作者 程序员写代码, 作者简介 Lee
TreeNode rightChild;intvalue;publicTreeNode(intvalue){this.value = value; }publicTreeNode(){ } }publicstaticvoidmain(String[] args){BinaryTreeInOrderTraversalbinaryTreeInOrderTraversal=newBinaryTreeInOrderTraversal();char[] arr1 =newchar[]{'1','#','2','3'};char[] arr2 =newchar[]{'1...
binary a. 1.【计算机,数学】二进制的 2.【术语】仅基于两个数字的,二元的,由两部分组成的 tree n. 树,木料,树状物 vt. 赶上树 Tree 树状结构使计算机知道如何执行的机器指令。 in tree 【计】 入树 binary decimal 二-十进制的 binary multiplier 二进乘法器 decimal to binary 十翻二 decimal...
The number of nodes in the tree is in the range [0, 100]. -100 <= Node.val <= 100 英文版地址 leetcode.com/problems/b 中文版描述 给定一个二叉树的根节点 root ,返回 它的中序 遍历。 示例1: 输入:root = [1,null,2,3] 输出:[1,3,2] 示例2: 输入:root = [] 输出:[] 示例3:...
94. Binary Tree Inorder Traversal 题目描述 Given a binary tree, return theinorder For example: Given binary tree [1,null,2,3], 1 \ 2 / 3 return [1,3,2]. Note: 思路 本题的目的是将一个二叉树结构进行中序遍历输出...
publicList<Integer>inorderTraversal3(TreeNoderoot){List<Integer>ans=newArrayList<>();TreeNodecur=root;while(cur!=null){//情况 1if(cur.left==null){ans.add(cur.val);cur=cur.right;}else{//找左子树最右边的节点TreeNodepre=cur.left;while(pre.right!=null&&pre.right!=cur){pre=pre.right;}...
Given a binary tree, return theinordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 1. 2. 3. 4. 5. return[1,3,2]. 1 /** 2 * Definition for binary tree 3 * struct TreeNode { ...
18 A and B are two nodes on a binary tree. In the in-order traversal, the condition for A before B is ( ). A. A is to the left of B B. A is to the right of B C. A is the ancestor of B D. A is a descendant of B ...
* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */classSolution{public List<Integer>postorderTraversal(TreeNode root){Deque<Integer>result=newLinkedList<>();if(root==null){returnnewA...