[LeetCode] Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / 
  9  20
    /  
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

思路一：用两个vector分层次记录值，时间复杂度O(n)，空间复杂度O(n)

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int> > levelOrder(TreeNode *root) {
13         vector<vector<int> > result;
14         if (root == nullptr) return result;
15         
16         queue<TreeNode *> current, next;
17         vector<int> level;
18         
19         current.push(root);
20         while (!current.empty()) {
21             while (!current.empty()) {
22                     TreeNode *p = current.front();
23                     current.pop();
24                     level.push_back(p->val);
25                     if (p->left != nullptr) next.push(p->left);
26                     if (p->right != nullptr) next.push(p->right);
27             }
28             result.push_back(level);
29             level.clear();
30             swap(current, next);
31         }
32         
33         return result;
34     }
35 };

思路二：BFS的思想。用一个queue，然后每次层结束时插入0到队列中。c
　　　　参考：https://leetcode.com/discuss/28762/c-solution-using-only-one-queue-use-a-marker-null
　

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int> > ret;
        if (root == NULL) return ret;
        
        vector<int> level;
        queue<TreeNode *> q;
        q.push(root);
        q.push(0);
        while (!q.empty()) {
            TreeNode *p = q.front();
            q.pop();
            if (p) {
                level.push_back(p->val);
                if (p->left)
                    q.push(p->left);
                if (p->right)
                    q.push(p->right);
            } else { 
                 ret.push_back(level);
                 level.clear();
                 
                 //当发现空指针时，要检查队列中是否还有节点
                 if (!q.empty())
                    q.push(0);
            }
        }
        
        return ret;
    }
};