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:…

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…

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? 解法一:递归 /** * De…

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…

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]. 代码如下: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode…

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…

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 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…