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;    }};