Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree [1,null,2,3], return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? 解析 嘻嘻,最近因为马上要面试,就回忆一下树的遍历,快排,最长公共子序列,八皇后之类的基...
* struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> inorderTraversal(TreeNode*root) { vector<int>ret; stack<TreeNode *>sta; TreeNode*p =root;while( !sta.empt...
* public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * }*/publicclassSolution {/***@paramroot: The root of binary tree. *@return: Inorder in ArrayList which contains node values.*/publicArrayList<Integer>inorderTraversal(TreeNode root) {//write ...
2.2 last.right 不为 null,说明之前已经访问过,第二次来到这里,表明当前子树遍历完成,保存 cur 的值,更新 cur = cur.right public List<Integer> inorderTraversal3(TreeNode root) { List<Integer> ans = new ArrayList<>(); TreeNode cur = root; while (cur != null) { //情况 1 if (cur.left ...
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...
94. Binary Tree Inorder Traversal 两种解法,递归和迭代 Memory分别是6.1和5.7,速度好像是一样的。。 Recursive class Solution { public: vector<int>res; void in_order(TreeNode* root) { if (!root)return; in_order(root->left); res.push_back(root->val);...
本题链接:Binary Tree Inorder Traversal 本题标签:Tree, Hash Table, Stack 本题难度: 方案1: 时间复杂度: 空间复杂度:
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...
Binary tree traversal: Preorder, Inorder, and Postorder In order to illustrate few of the binary tree traversals, let us consider the below binary tree: Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the le...
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right (right) {} * }; */ class Solution { public: vector<int> inorderTraversal(TreeNode* root) { } }; 已存储 行1,列 1 运行和提交...