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].
思路：递归和迭代。
//递归
class Solution {
public:
void help(vector<int>& res, TreeNode* root)
{
if (!root) return;
res.push_back(root->val);
help(res, root->left);
help(res, root->right);
}
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
help(res, root);
return res;
}
};
//迭代
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
if (!root) return res;
stack<TreeNode *> sta;
sta.push(root);
while (!sta.empty())
{
TreeNode* cur = sta.top();
sta.pop();
res.push_back(cur->val);
if (cur->right) sta.push(cur->right);
if (cur->left) sta.push(cur->left);
}
return res;
}
};
Binary Tree Preorder Traversal -- LeetCode
原文：http://www.cnblogs.com/fenshen371/p/5167937.html