Binary Tree Preorder Traversal and Binary Tree Postorder Traversal

Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

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

   1
    \
     2
    /
   3

return [1,2,3].

c++版：

/**
 * 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> preorderTraversal(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<int> result;
        vector<int> left;
        vector<int> right;  

        if(root == NULL) return result;
        result.push_back(root->val);
        left = preorderTraversal(root->left);
        right = preorderTraversal(root->right);  

        if(left.size() != )
            result.insert(result.end(), left.begin(), left.end());
        if(right.size() != )
            result.insert(result.end(), right.begin(), right.end());  

        return result;  

    }
};

Binary Tree Postorder Traversal

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].

C++版本：

/**
 * 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> postorderTraversal(TreeNode *root) {
        vector<int> ret;
        dfs(root, ret);
        return ret;
    }  

    void dfs(TreeNode* root, vector<int>& ret)
    {
        if(NULL == root)
            return ;
        dfs(root->left, ret);
        dfs(root->right, ret);
        ret.push_back(root->val);
    }     

};