題目：給出一棵二叉樹，返回其節點值的後序遍歷。

給出一棵二叉樹 {1,#,2,3},

   1
         2
    /
   3

返回 [3,2,1]

你能使用非遞歸實現麽？

解：遞歸解，非遞歸以後補充

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 
 
*/


class Solution {
public:
    /*
     * @param root: A Tree
     * @return: Postorder in ArrayList which contains node values.
     */
    vector<int> postorderTraversal(TreeNode * root) {
        // write your code here
        vector<int> re;
        if(root!=NULL)
        {
            postorder(root,re);
        }
        
 
return re;
    }
    void postorder(TreeNode* node,vector<int> &re)
    {
        if(node==NULL)
            return;
        if(node->left!=NULL)
            postorder(node->left,re);
        if(node->right!=NULL)
            postorder(node->right,re);
        re.push_back(node
 
->val);
    }
};

LintCode 68. 二叉樹的後序遍歷