Leetcode-Binary Tree Level Order Traversal II

 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>> levelOrderBottom(TreeNode root) {
12         List<List<Integer>> res = new ArrayList<List<Integer>>();
13         if (root==null)
14             return res;
15         
16         //Visit all node by using BFS.
17         List<TreeNode> queue = new ArrayList<TreeNode>();
18         List<Integer> depth = new ArrayList<Integer>();
19         queue.add(root);
20         depth.add(1);
21         int index = 0;
22         int curDepth = -1;
23         TreeNode curNode = null;
24         while (index<queue.size()){
25             curNode = queue.get(index);
26             curDepth = depth.get(index);
27             if (curNode.left!=null){
28                 queue.add(curNode.left);
29                 depth.add(curDepth+1);
30             }
31             
32             if (curNode.right!=null){
33                 queue.add(curNode.right);
34                 depth.add(curDepth+1);
35             }
36             
37             index++;
38         }
39         
40         //Get the max depth, which is the number of lists in the result.
41         int maxDepth = depth.get(depth.size()-1);
42         for (int i=0;i<maxDepth;i++)
43             res.add(new ArrayList<Integer>());
44         
45         //Put the value of each node into the corresponding list in the result.
46         for (int i=0;i<queue.size();i++){
47             curNode = queue.get(i);
48             curDepth = depth.get(i);
49             index = maxDepth-curDepth;
50             List<Integer> tempList = res.get(index);
51             tempList.add(curNode.val);
52         }
53         
54         return res;
55     }
56 }