[LeetCode] Binary Tree Preorder Traversal [递归版]

题目：

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]

.

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

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
vector<int> result;
public:
vector<int> preorderTraversal(TreeNode *root) {
return tmpFuction(root, result);
}

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

如果采用递归方式来做，思路是很简单的。关键是返回值result的处理：因为不能在preorderTraversal中定义vector<int>，所以定义了临时函数，把result作为一个入口参数传入其中并进行操作。