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[Leetcode] Binary Tree Level Order Traversal

时间:2014-11-02 07:01:30      阅读:210      评论:0      收藏:0      [点我收藏+]

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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]
]

Solution 1: BFS

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<List<Integer>> levelOrder(TreeNode root) {
12         List<List<Integer>> ret=new ArrayList<List<Integer>>();
13         if(root==null)
14             return ret;
15         
16         Queue<TreeNode> queue=new LinkedList<TreeNode>();
17         List<Integer> al=new ArrayList<Integer>();
18         
19         queue.add(root);
20         int curLvl=1;
21         int nexLvl=0;
22         
23         while(!queue.isEmpty()){
24             TreeNode cur=queue.remove();
25             al.add(cur.val);
26             curLvl--;
27             
28             if(cur.left!=null){
29                 queue.add(cur.left);
30                 nexLvl++;
31             }
32             if(cur.right!=null){
33                 queue.add(cur.right);
34                 nexLvl++;
35             }
36             if(curLvl==0){
37                 ret.add(al);
38                 al=new ArrayList<Integer>();
39                 curLvl=nexLvl;
40                 nexLvl=0;
41             }
42         }
43         
44         return ret;
45     }
46 }

 

Solution 2: DFS

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     
12     List<List<Integer>> ret;
13     
14     public List<List<Integer>> levelOrder(TreeNode root) {
15         ret = new ArrayList<List<Integer>>();
16         goDeeper(0, root);
17         return ret;
18     }
19 
20     private void goDeeper(int level, TreeNode root) {
21         // TODO Auto-generated method stub
22         if (root == null)
23             return;
24         if (ret.size() > level) {
25             List<Integer> a = ret.get(level);
26             a.add(root.val);
27         } else {
28             List<Integer> a = new LinkedList<Integer>();
29             a.add(root.val);
30             ret.add(a);
31         }
32         goDeeper(level+1, root.left);
33         goDeeper(level+1, root.right);
34     }
35 }

 

[Leetcode] Binary Tree Level Order Traversal

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原文地址:http://www.cnblogs.com/Phoebe815/p/4068305.html

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