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

For example:

Given binary tree {1,#,2,3},



return [1,2,3].

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

先序遍历应该是三种遍历方式中最好使用非递归来实现的了。

能够每次訪问一个节点后再将其右子节点入栈，然后左子节点入栈。然后再从栈中取元素，能够发现就是先序遍历的结果。

须要注意的是假设子节点不存在就不须要进行入栈操作。

runtime:4ms

/**

 * 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(root==NULL)

            return res;

        stack<TreeNode *> s;

        TreeNode * node=root;

        s.push(node);

        while(!s.empty())

        {

            node=s.top();

            s.pop();

            res.push_back(node->val);

            if(node->right)  s.push(node->right);

            if(node->left)  s.push(node->left);

        }

        return res;

    }

};