题目：

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.

思路：

经典的中序排序算法，可以用递归实现，也可以用带权重的堆栈实现。

代码：

/**
 * 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<int> inorderTraversal(TreeNode *root) {
        if(root == NULL){
            return vector<int>();
        }
        
        vector<int> left = inorderTraversal(root->left);
        left.push_back(root->val);
        vector<int> right = inorderTraversal(root->right);
        for(int i = 0; i < right.size(); i++){
            left.push_back(right[i]);
        }
        return left;
    }
};