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?

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.

/**
 * Definition for a binary tree node.
 * 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) {
        vector<int> result;
        if(!root) return result;
        
        inorder(root, result);
        return result;
    }
    void inorder(TreeNode* root, vector<int>& result){
        if(root->left) inorder(root->left, result);
        result.push_back(root->val);
        if(root->right) inorder(root->right, result);
    }
};

2. Iteration

    Runtime: 0ms.

/**
 * Definition for a binary tree node.
 * 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) {
        vector<int> result;
        if(!root) return result;
        stack<TreeNode* > stk;
        
        while(root || !stk.empty()){
            if(root){
                stk.push(root);
                root = root->left;
            }
            else{
                root = stk.top();
                result.push_back(root->val);
                stk.pop();
                root = root->right;
            }
        }
        return result;
    }
};