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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.
Tree Hash Table Stack
#include <iostream> #include <vector> using namespace std; /** * 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 > ret; vector<int> inorderTraversal(TreeNode *root) { ret.clear(); helpFun(root); return ret; } void helpFun(TreeNode * node) { if(node ==NULL) return ; helpFun(node->left); ret.push_back(node->val); helpFun(node->right); } }; int main() { return 0; }
[LeetCode] Binary Tree Inorder Traversal 中序排序
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原文地址：http://www.cnblogs.com/Azhu/p/4346002.html
