Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree [1,null,2,3], 1 \ 2 / 3 return [1,3,2]. Note: Recursive(递归) solution is trivial, could you do it iteratively(迭代)? 思路: 解法一:用递归方法很简单, (1)如果root为空,则返回…

给定一个二叉树,返回其中序遍历.例如:给定二叉树 [1,null,2,3],   1    \     2    /   3返回 [1,3,2].说明: 递归算法很简单,你可以通过迭代算法完成吗?详见:https://leetcode.com/problems/binary-tree-inorder-traversal/description/ Java实现: 递归实现: /** * Definition for a binary tree node. * public class TreeNo…

http://oj.leetcode.com/problems/binary-tree-inorder-traversal/ 树的中序遍历,递归方法,和非递归方法. /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ cla…

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what "{1,#,2,3}" means? > re…

题目大意 https://leetcode.com/problems/binary-tree-inorder-traversal/description/ 94. Binary Tree Inorder Traversal Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up:…

94. Binary Tree Inorder Traversal    二叉树的中序遍历 递归方法: 非递归:要借助栈,可以利用C++的stack…

1.  Binary Tree Inorder Traversal Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? 分析:求二叉树的…

Binary Tree Inorder Traversal Total Accepted: 16406 Total Submissions: 47212My Submissions Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. public class Solutio…

Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what "{1,#,2,3}" means? > r…

Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [,,] \ / Output: [,,] Follow up: Recursive solution is trivial, could you do it iteratively? 题目中要求使用迭代用法,利用栈的“先进后出”特性来实现中序遍历. 解法一:(迭代)将根节点压入栈,当其左子树存在时,一直将其左子树压入栈,…

Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively? 题意: 二叉树中序遍历 Solution1:   Recursion code class Soluti…

这道题是LeetCode里的第94道题. 题目要求: 给定一个二叉树,返回它的中序 遍历. 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出: [1,3,2] 进阶: 递归算法很简单,你可以通过迭代算法完成吗? 解题代码: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) :…

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree [1,null,2,3], 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? 求二叉树的中序遍历,要求不是用递归. 先用递归做一下,很简单. /** * Defi…

给定树根root.实现中序遍历,也就是左根右. 用递归的话,很简单,左边的返回值加上root的再加上右边的就行. 我自己写的有点挫: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { pu…

二叉树的中序遍历,即左子树,根, 右子树 /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void dfs(vector<int> &ans,TreeNode…

点击打开题目链接 今天只是写了递归的版本,因为还没想好怎么用迭代来实现,可以写的过程中,有一点是有疑问的,虽然我的代码可以AC. 问题是:主调函数是可以使用子函数中返回的在子函数中定义的vector. 我认为在被调函数执行结束之后,其分配的空间是应该被释放的,所以在被调函数中定义的变量也是不可以被主调函数中使用的...可能是C++的基础知识又给忘了,有空要拾起来了. 附上代码(递归版): /** * Definition for binary tree * struct TreeNode { *…

树的中序遍历.先不断压入左结点至末尾,再访问,再压入右结点.注意和先序遍历的比较 vector<int> inorderTraversal(TreeNode *root) { vector<int> result; stack<TreeNode *>s; TreeNode *p = root; while (!s.empty() || p != nullptr) { if (p != nullptr) { //将当前结点和其左结点不断入栈 s.push(p); p =…

解题思路: 中序遍历,左子树-根节点-右子树 JAVA实现如下: public List<Integer> inorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<Integer>(); if(root==null) return list; if (root.left != null) list.addAll(inorderTraversal(root.left)); list.add(…

非递归的中序遍历,要用到一个stack class Solution { public: vector<int> inorderTraversal(TreeNode* root) { vector<int> ret; if(!root) return ret; //a(ret) stack<TreeNode*> stk; stk.push(root); //ahd(root) //a(stk) //dsp TreeNode* p=root; while(p->le…

二叉树遍历(前序.中序.后序.层次.深度优先.广度优先遍历) 描述 解析 递归方案 很简单,先左孩子,输出根,再右孩子. 非递归方案 因为访问左孩子后要访问右孩子,所以需要栈这样的数据结构. 1.指针指向根,根入栈,指针指向左孩子.把左孩子当作子树的根,继续前面的操作. 2.如果某个节点的左孩子不存在,节点出栈,指针指向节点的右孩子.把这个右节点当作根, 继续前面的操作. 代码 /** * Definition for a binary tree node. * public class Tre…

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree [1,null,2,3], 1 \ 2 / 3 return [1,3,2]. 本题如果用recursive的方法非常简单.这里主要考察用iterative的方法求解.例如: [1,3,4,6,7,8,10,13,14] 从8开始依次将8,3,1 push入栈.这时root=None,每当roo…

题目描述: Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. 解题思路: 使用栈.从根节点开始迭代循环访问,将节点入栈,并循环将左子树入栈.如果当前节点为空,则弹出栈顶节点,也就是当前节点的父节点,并将父节点的值加入到list中,然后选择右节点作为循环的节点,依次循环.…

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3} 1 \ 2 / 3 return [1,2,3]. 前序遍历二叉树,只不过题目的要求是尽量不要使用递归,当然我还是先用递归写了一个: /** * Definition for a binary tree node. * struct TreeNode { * int val…

方法一:(递归) class Solution { public: vector<int> inorderTraversal(TreeNode* root) { vector<int> v; inorderTraversalHelp(root, v); return v; } void inorderTraversalHelp(TreeNode *root, vector<int>& v) { if(root) { inorderTraversalHelp(ro…

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 树的遍历,最常见的有先序遍历,中序遍历,后序遍历和层序遍历,它们用递归实现起来都非常的简…

翻译 给定一个二叉树.返回其兴许遍历的节点的值. 比如: 给定二叉树为 {1. #, 2, 3} 1 \ 2 / 3 返回 [3, 2, 1] 备注:用递归是微不足道的,你能够用迭代来完毕它吗? 原文 Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: R…

一.题目说明 题目94. Binary Tree Inorder Traversal,给一个二叉树,返回中序遍历序列.题目难度是Medium! 二.我的解答 用递归遍历,学过数据结构的应该都可以实现. class Solution{ public: vector<int> inorderTraversal(TreeNode* root){ if(root != NULL){ if(root->left !=NULL)inorderTraversal(root->left); res…

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively?中序遍历二叉树,递归遍历当然很容易,题目还要求不用递归,下面给出两种方法: 递归: /**…

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what "{1,#,2,3}" means? > re…

既上篇关于二叉搜索树的文章后,这篇文章介绍一种针对二叉树的新的中序遍历方式,它的特点是不需要递归或者使用栈,而是纯粹使用循环的方式,完成中序遍历. 线索二叉树介绍 首先我们引入“线索二叉树”的概念: "A binary tree is threaded by making all right child pointers that would normally be null point to the inorder successor of the node, and all left chi…