94. 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    /   3return [1,3,2].Note: Recursive solution is trivial, could you do it iteratively?借助栈进行迭代来模拟函数栈*//** * 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;        if(!root) return res;        stack<TreeNode*> s;        TreeNode* p=root;        while(!s.empty() || p)        {            if(p)            {                s.push(p);                p=p->left;            }            else            {                TreeNode* pNode=s.top();                res.push_back(pNode->val);                s.pop();                p=pNode->right;            }        }        return res;    }};