Binary Tree Level Order Traversal II

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

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

    3

   / \

  9  20

    /  \

   15   7

 

return its bottom-up level order traversal as:

[

  [15,7]

  [9,20],

  [3],

]
解题思路： 使用BFS遍历树，将每一层放入一个ArrayList中。当一层结束后，将这个ArrayList插到最终的List的头部。

 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     public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {

13         // Start typing your Java solution below

14         // DO NOT write main() function

15         ArrayList<TreeNode> queue = new ArrayList<TreeNode>();

16         ArrayList<ArrayList<Integer>> r = new ArrayList<ArrayList<Integer>>();

17         if(root == null) return r;

18         queue.add(root);

19         TreeNode nl = new TreeNode(999);//999 是一个singal，表示一层结束了。

20         queue.add(nl);

21         ArrayList<Integer> cur = new ArrayList<Integer>();

22         while(queue.size() != 1){

23             TreeNode rn = queue.remove(0);

24             if(rn.val == 999){

25                 queue.add(nl);

26                 r.add(0,cur);

27                 cur = new ArrayList<Integer>();

28             }

29             else{

30                 if(rn.left != null) queue.add(rn.left);

31                 if(rn.right != null) queue.add(rn.right);

32                 cur.add(rn.val);

33             }

34         }

35         r.add(0,cur);

36         return r;

37     }

38 }