LeetCode 144 Binary Tree Preorder Traversal（二叉树前序遍历）

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?

题目大意：写出二叉树的非递归前序遍历的方法。

解题思路：和中序遍历的那一道题类似，处理过程如下：

    对于任意节点N：

    1. 访问N并将N入栈

    2. 判断结点N的左孩子是否为空，若为空，则取栈顶结点并进行出栈操作，并将栈顶结点的右孩子置为当前结点N，循环至1;若不为空，则将N的左孩子置为当前结点N

    3. 直到N为空且栈也为空，遍历结束

参考资料：http://www.cnblogs.com/dolphin0520/archive/2011/08/25/2153720.html

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) {
if(root == nullptr) return vector<int>();

stack<TreeNode*> stk;
vector<int> ans;
TreeNode* now = root;

while(now || stk.size()){
while(now){
stk.push(now);
ans.push_back(now->val);
now = now->left;
}

if(stk.size()){
now = stk.top();
stk.pop();
now = now->right;
}
}
return ans;
}
};