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…

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…

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…

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. 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…

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…

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…