144. Binary Tree Preorder Traversal

题目：Binary Tree Preorder Traversal

原题链接：https://leetcode.com/problems/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?

给出二叉树先序遍历的结果，尝试用递归和非递归的方法。

递归：按照二叉树先序遍历的顺序访问即可：

/**
* 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:
void preOrder(TreeNode* root, vector<int>& ans) {
if(root == NULL) return;
ans.push_back(root -> val);
preOrder(root -> left, ans);
preOrder(root -> right, ans);
}
vector<int> preorderTraversal(TreeNode* root) {
vector<int> ans;
preOrder(root, ans);
return ans;
}
};

非递归：需要用到栈（stack）,由于栈的特性是后进显出（LIFO，Last In First Out），所以在子树压栈的时候需要先压右子树再压左子树，代码如下：

/**
* 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> preorderTraversal(TreeNode* root) {
vector<int> ans;
stack<TreeNode* > st;
if(root) st.push(root);
while(!st.empty()) {
TreeNode* front = st.top();
ans.push_back(front -> val);
st.pop();
if(front -> right) st.push(front -> right);
if(front -> left) st.push(front -> left);
}
return ans;
}
};