Binary Tree Preorder Traversal--二叉树的先序遍历

原题：

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

=>给出一个二叉树，返回先序遍历的所有的节点值

For example:

例如：
Given binary tree {1,#,2,3},

给出下面的二叉树

   1
\
2
/
3

return [1,2,3].

返回[1,2,3]

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

=》注意：递归的算法是很普通，能不能不要递归呢？

/**
* 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.
}
};

晓东解析：

这个题目用递归来实现的确很简单，就是先遍历左，再遍历右。那非递归的算法就是先一直左到底，并把这些所有的左都压到一个栈里面，然后到最后，再一个个pop出来，每pop一个，对它的右子树再进行一次类似的遍历就可以了。

代码实现：

1、递归实现

/**
* 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() != 0)
result.insert(result.end(), left.begin(), left.end());
if(right.size() != 0)
result.insert(result.end(), right.begin(), right.end());
return result;
}
};

运行结果：

2、非递归实现：