Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {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.

Analyse: root->left, root, root->right.

1. Recursion

Runtime: 0ms.

 1 /**
 2  * Definition for a binary tree node.
 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         vector<int> result;
14         if(!root) return result;
15         
16         inorder(root, result);
17         return result;
18     }
19     void inorder(TreeNode* root, vector<int>& result){
20         if(root->left) inorder(root->left, result);
21         result.push_back(root->val);
22         if(root->right) inorder(root->right, result);
23     }
24 };

2. Iteration

Runtime: 0ms.

 1 /**
 2  * Definition for a binary tree node.
 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         vector<int> result;
14         if(!root) return result;
15         stack<TreeNode* > stk;
16         
17         while(root || !stk.empty()){
18             if(root){
19                 stk.push(root);
20                 root = root->left;
21             }
22             else{
23                 root = stk.top();
24                 result.push_back(root->val);
25                 stk.pop();
26                 root = root->right;
27             }
28         }
29         return result;
30     }
31 };