Binary Tree Inorder Traversal

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

For example:
Given binary tree [1,null,2,3],

   1    \     2    /   3

return [1,3,2].

class Solution {public:    vector<int> inorderTraversal(TreeNode* root) {        vector<int> result;        if(root != NULL)            //recursiveInorderTraversal(root, result);        return result;    }private:    /*    void recursiveInorderTraversal(TreeNode* root, vector<int> &result)    {        if(root == NULL)            return;        if(root->left != NULL)            recursiveInorderTraversal(root->left, result);        result.push_back(root->val);        if(root->right != NULL)            recursiveInorderTraversal(root->right, result);    }    */    void iterativelInorderTraversal(TreeNode* root, vector<int> &result)    {        stack<TreeNode *> leftSubTree;        TreeNode *p = root;        while(p != NULL || !leftSubTree.empty())        {            while(p != NULL) //将左子树一直入栈            {                leftSubTree.push(p);                p = p->left;            }            if(!leftSubTree.empty())            {                p = leftSubTree.top();                result.push_back(p->val);                leftSubTree.pop();                p = p->right; //开始遍历右子树            }        }    }};