Given a binary tree, return the postorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

二叉树的后序遍历

递归算法：

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int>res;
        if(root==NULL) return res;
        return postorder(root,res);
    }
    vector<int> postorder(TreeNode* root,vector<int>& res)
    {
        if(root)
        {
            postorder(root->left,res);
            postorder(root->right,res);
            res.push_back(root->val);
        }
        return res;
    }
};

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int>res;
        if(root==NULL) return res;
        stack<TreeNode*> qu;
        TreeNode* node=root;
        while(!qu.empty()||node)
        {
            if(node)
            {
                qu.push(node);
                res.push_back(node->val);
                node=node->right;
            }
            else
            {
                TreeNode* cur=qu.top();
                qu.pop();
                node=cur->left;
            }
        }
        reverse(res.begin(),res.end());
        return res;
    }
};