[leetcode][tree] Binary Tree Preorder Traversal

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]

.
递归实现：

/**
* 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:
//最直观的想法：递归
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
preorderTraversalCore(root, res);
return res;
}
private:
void preorderTraversalCore(TreeNode *root, vector<int> &res){
if(NULL == root) return;
res.push_back(root->val);
preorderTraversalCore(root->left, res);
preorderTraversalCore(root->right, res);
}
};

非递归实现：

/**
* 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> res;
if (NULL == root) return res;
stack<TreeNode *> sta;
TreeNode  *p = root;
while (p || !sta.empty()){
while (p){//一直向左
res.push_back(p->val);//访问节点
sta.push(p);//节点入栈
p = p->left;//继续向左
}
p = sta.top();
sta.pop();//弹出栈顶节点
p = p->right;//更新p为栈顶节点的右孩子，开始下一轮循环
}
return res;
}
};