145. Binary Tree Postorder Traversal[hard]

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?

水题，输出二叉树的后序

/** * 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> ans;    void dfs(TreeNode* root)    {        if(root == NULL)            return;        if(root->left != NULL)            postorderTraversal(root->left);        if(root->right != NULL)            postorderTraversal(root->right);        ans.push_back(root->val);    }    vector<int> postorderTraversal(TreeNode* root) {        dfs(root);        return ans;    }    };