本网页所有文字内容由 imapbox邮箱云存储,邮箱网盘, iurlBox网页地址收藏管理器 下载并得到。
ImapBox 邮箱网盘 工具地址: https://www.imapbox.com/download/ImapBox.5.5.1_Build20141205_CHS_Bit32.exe
PC6下载站地址：PC6下载站分流下载
本网页所有视频内容由 imoviebox边看边下-网页视频下载, iurlBox网页地址收藏管理器 下载并得到。
ImovieBox 网页视频 工具地址: https://www.imapbox.com/download/ImovieBox4.7.0_Build20141115_CHS.exe
本文章由: imapbox邮箱云存储,邮箱网盘,ImageBox 图片批量下载器,网页图片批量下载专家,网页图片批量下载器,获取到文章图片,imoviebox网页视频批量下载器,下载视频内容,为您提供.
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 more on how binary tree is serialized on OJ.
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> inorderTraversal(TreeNode *root) { stack<TreeNode *> nodestack; vector<int> arr(0); TreeNode *node; if(root == NULL) return arr; node = root; while(!nodestack.empty() || node!=NULL){ if(node!=NULL) { nodestack.push(node); node=node->left; } else{ node=nodestack.top(); arr.push_back(node->val); nodestack.pop(); node=node->right; } } return arr; } };
阅读和此文章类似的: 程序员专区
官方软件产品操作指南 (170)