[leedcode 107] 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]
]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        //跟正常的层序遍历的不同，在放入结果时，头插法
        List<List<Integer>> res=new ArrayList<List<Integer>>();
        List<Integer> seq=new ArrayList<Integer>();
        if(root==null) return res;
        int curLevel=1;
        int nextLevel=0;
        LinkedList<TreeNode> queue=new LinkedList<TreeNode>();//如果是List，有问题，会显示remove方法找不到
        queue.add(root);
        while(!queue.isEmpty()){
            TreeNode node=queue.remove();
            seq.add(node.val);
            curLevel--;
            if(node.left!=null){
                queue.add(node.left);
                nextLevel++;
            }
            if(node.right!=null){
                queue.add(node.right);
                nextLevel++;
            }
            if(curLevel==0){
                res.add(0,seq);
                curLevel=nextLevel;
                nextLevel=0;
                seq=new ArrayList<Integer>();
            }
        }
        return res;
    }
}