55. Binary Tree Preorder Traversal

1. Binary Tree Preorder Traversal My Submissions QuestionEditorial Solution
   Total Accepted: 119655 Total Submissions: 300079 Difficulty: Medium
   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].

思路：用栈来进行先序遍历，先入栈的后遍历
时间复杂度：O(n)
空间复杂度：O(1)

/**
 * 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> vtn;
        if(root==NULL)return vtn;  //为空时先处理，不然后面死循环
        stack<TreeNode*> stn;
        stn.push(root);
        while(!stn.empty()){
            TreeNode *p = stn.top();
            vtn.push_back(p->val);
            stn.pop();
            if(p->right!=NULL)stn.push(p->right);
            if(p->left!=NULL)stn.push(p->left);
        }
        return vtn;
    }
};