LeetCode之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?

#include <stack>
#include <vector>
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

// 不断往左走并访问，直到无法走后，回溯到先一个结点的右边，再重复
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {

stack <TreeNode *> intStack;
vector<int> intVector;
TreeNode *p = root;
do{
while (p){
intVector.push_back(p->val);
intStack.push(p);
p = p->left;
}//while
if(!intStack.empty()){
p = intStack.top()->right;
intStack.pop();
}
} while (!intStack.empty()||p);
return intVector;
}
};

//用栈，采取先放右子树再放左子树的方法，达到输出时先左后右
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {

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