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…

题目意思:二叉树中序遍历,结果存在vector<int>中 解题思路:迭代 迭代实现: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector&l…

题目:Binary Tree Inorder Traversal 二叉树的中序遍历,和前序.中序一样的处理方式,代码见下: struct TreeNode { int val; TreeNode* left; TreeNode* right; TreeNode(int x): val(x), left(NULL),right(NULL) {} }; vector<int> preorderTraversal(TreeNode *root) //非递归的中序遍历(用栈实现) { if (NULL…

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? > read m…

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:…

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. 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: /**…

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…

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…

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…

题目:Binary Tree Postorder Traversal 二叉树的后序遍历,题目要求是采用非递归的方式,这个在上数据结构的课时已经很清楚了,二叉树的非递归遍历不管采用何种方式,都需要用到栈结构作为中转,代码很简单,见下: struct TreeNode { int val; TreeNode* left; TreeNode* right; TreeNode(int x): val(x), left(NULL),right(NULL) {} }; vector<int> preord…

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…

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…

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…

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…

144. Binary Tree Preorder Traversal Difficulty: Medium 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 it…

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: Recursive solution is trivial, could you do it iteratively? 后序遍历:左孩子->右孩子->根节点 后序遍历最关键的是利用一个指针保存前一个访问过的信…

这道题是LeetCode里的第144道题. 题目要求: 给定一个二叉树,返回它的 前序 遍历. 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出: [1,2,3] 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)…

题目链接 题目大意:中序遍历二叉树.先序见144,后序见145. 法一:DFS,没啥说的,就是模板DFS.代码如下(耗时1ms): public List<Integer> inorderTraversal(TreeNode root) { List<Integer> res = new ArrayList<Integer>(); dfs(res, root); return res; } private void dfs(List<Integer> res…

题目意思:二叉树先序遍历,结果存在vector<int>中 解题思路:1.递归(题目中说用递归做没什么意义,我也就贴贴代码吧) 2.迭代 迭代实现: class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> ans; if(root){ TreeNode* temp; stack<TreeNode*> s; //利用栈,每次打印栈顶,然后将栈顶弹出…

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. Example: Input: [,,] \ / Output: [,,] Follow up: Recursive solution is trivial, could you do it iteratively? 方法一:使用迭代(C++) vector<int> preorderTraversal(TreeNode* root) { vecto…

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…

144. Binary Tree Preorder Traversal 前序的非递归遍历:用堆来实现 如果把这个代码改成先向堆存储左节点再存储右节点,就变成了每一行从右向左打印 如果用队列替代堆,并且先存储左节点,再存储右节点,就变成了逐行打印 class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> result; if(root == NULL) return res…

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…

题目:Binary Tree Preorder Traversal 二叉树的前序遍历,同样使用栈来解,代码如下: struct TreeNode { int val; TreeNode* left; TreeNode* right; TreeNode(int x): val(x), left(NULL),right(NULL) {} }; vector<int> preorderTraversal(TreeNode *root) //非递归的前序遍历(用栈实现) { if (NULL == r…

题目: 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? OJ's Binary Tree Serialization: The seria…

题目描述 (原题目链接) 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]. 解题思路 这道题目是关于二叉树中序遍历的迭代实现.之前就总结过二叉树的非递归实现.可是做题的时候太久没刷题,就断路了.所以重新思考了一种解法.主要想法是:用一个标记位标记是否需要遍历当前节点…