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].

Note: Recursive solution is trivial, could you do it iteratively?
Tree Stack
分别用递归和栈来解决。

递归：

    vector<int> result;    void preorder(TreeNode* p){        if (p == NULL)            return;        result.push_back(p->val);        preorder(p->left);        preorder(p->right);    }    vector<int> preorderTraversal(TreeNode* root) {        preorder(root);        return result;//隔开，才能返回    }

栈：（空间复杂度O（n））

vector<int> preorderTraversal(TreeNode* root) {        vector<int> arr;        stack<TreeNode*> s;        TreeNode* p;        if (root == NULL)            return arr;        s.push(root);        while (!s.empty()){            p = s.top();            arr.push_back(p->val);            s.pop();            if (p->right)                s.push(p->right);            if (p->left)                s.push(p->left);        }        return arr;}