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。

/**
 * 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>> levelOrderBottom(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<TreeNode> queue = new ArrayList<TreeNode>();
        ArrayList<ArrayList<Integer>> r = new ArrayList<ArrayList<Integer>>();
        if(root == null) return r;
        queue.add(root);
        TreeNode nl = new TreeNode(999);//999 是一个singal，表示一层结束了。
        queue.add(nl);
        ArrayList<Integer> cur = new ArrayList<Integer>();
        while(queue.size() != 1){
            TreeNode rn = queue.remove(0);
            if(rn.val == 999){
                queue.add(nl);
                r.add(0,cur);
                cur = new ArrayList<Integer>();
            }
            else{
                if(rn.left != null) queue.add(rn.left);
                if(rn.right != null) queue.add(rn.right);
                cur.add(rn.val);
            }
        }
        r.add(0,cur);
        return r;
    }
}