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?

非递归中序遍历，基础吧， s.pop（） 原来是返回空的。。。

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector
    
      inorderTraversal(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        stack
     
       S;    	vector
      
        ret;		ret.clear();		if(root==NULL)			return ret;		TreeNode *p =root; 		while(!S.empty()|| p)		{			if(p!=NULL)			{				S.push(p);				p= p->left;			}else{				p = S.top();				S.pop();				ret.push_back(p->val);				p = p->right;			}		}		return ret;    }};