[Tree]Binary Tree Inorder Traversal

Total Accepted: 98729 Total Submissions: 261539 Difficulty: Medium

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?

 (M) Validate Binary Search Tree (M) Binary Tree Preorder Traversal (H) Binary Tree Postorder Traversal (M) Binary Search Tree Iterator(M) Kth Smallest Element in a BST (H) Closest Binary Search Tree Value II (M) Inorder Successor in BST

/**
 * 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> res;
        stack<TreeNode*> stk;
        while(root || !stk.empty()){
            while(root){
                stk.push(root);
                root=root->left;
            }
            if(!stk.empty()){
                root = stk.top();
                stk.pop();
                res.push_back(root->val);
                root = root->right;
            }
        }
        return res;
    }
};

Next challenges: (M) Validate Binary Search Tree (H) Binary Tree Postorder Traversal (M) Binary Search Tree Iterator (M) Kth Smallest Element in a BST (H) Closest Binary Search Tree Value II(M) Inorder Successor in BST