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?
The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.
Here‘s an example:
1
/ 2 3
/
4
5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> inorderTraversal(TreeNode *root) { 13 14 vector<int> result; 15 inOrder(root, result); 16 return result; 17 18 } 19 20 void inOrder(TreeNode *root, vector<int> &result) 21 { 22 if(root != NULL) 23 { 24 inOrder(root->left, result); 25 result.push_back(root->val); 26 inOrder(root->right, result); 27 } 28 } 29 };
原文：http://www.cnblogs.com/YQCblog/p/3970200.html