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 { *…
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…
Binary Tree Zigzag Level Order Traversal Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree {3,9,20,…
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    /   3return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? co…
Binary Tree Inorder Traversal Given a binary tree, return the inorder traversal of its nodes' values. Example Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. For in-order traversal we all know that firstly we want to go to the lowest level and…
原题: Binary Tree Inorder Traversal 和 3月3日(2) Binary Tree Preorder Traversal 类似,只不过变成中序遍历,把前序遍历的代码拿出来,改函数,改一句话位置 AC.…
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:…
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…
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…