LeetCode-Binary Tree Level Order Traversal

Binary Tree Level Order Traversal

##题目

####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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

####OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:

  1
 / \
2   3
   /
  4
   \
    5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

##解题思路
该题是对二叉树进行层次优先遍历，层次遍历主要采用队列的形式进行存储，通过将每个节点的左孩子和右孩子放入队列中，然后每次从队列中取出元素即可。在java中使用LinkedList来实现队列操作，其中主要方法为：入队：addLast(),出队：revomeFirst(),获取第一个元素：getFirst(),判断队列是否为空：isEmpty();

##算法代码
代码采用JAVA实现：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> res=new ArrayList<ArrayList<Integer>>();
        //该队列中放入节点
        LinkedList<TreeNode> queen=new LinkedList<TreeNode>();
        //该队列中放入queen中对应节点的层次
        LinkedList<Integer> cequeen=new LinkedList<Integer>();
        if(root==null)
        	return res;
        queen.addLast(root);
        cequeen.addLast(0);
        int cengshu=0;
        ArrayList<Integer> cur=new ArrayList<Integer>();
        while(!queen.isEmpty())
        {
        	int ce=cequeen.getFirst();
        	if(ce==cengshu)
        	{
        		TreeNode no=queen.removeFirst();
        		int noce=cequeen.removeFirst();
        		cur.add(no.val);
        		if(no.left!=null)
        		{
        			queen.addLast(no.left);
        			cequeen.addLast(noce+1);
        		} 			
        		if(no.right!=null)
        		{
        			queen.addLast(no.right);
        			cequeen.addLast(noce+1);
        		}       			
        	}else{
        		res.add(new ArrayList<Integer>(cur));
        		cur.clear();
        		cengshu++;
        	}
        }
        res.add(new ArrayList<Integer>(cur));
        return res;
    }
}

Comments

▲