Leetcode-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]
]

分析：在層次遍歷的基礎上增加了對結果的一個翻轉。

AC代碼：

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> answerList = new ArrayList<List<Integer>>();
		Queue<TreeNode> q = new LinkedList<TreeNode>();
		TreeNode nextLevelStarter;
		q.add(root);
		if( root == null )
            return answerList;
		while( !q.isEmpty() ){
			TreeNode currentNode = q.peek();
			nextLevelStarter = null;
			List<Integer> thisLevelList = new ArrayList<Integer>();
			//find nextLevelStart
			while( q.peek() != nextLevelStarter && q.isEmpty() == false ){
				currentNode = q.remove();
				if( nextLevelStarter == null ){
					if( currentNode.left != null )
						nextLevelStarter = currentNode.left;
					else if( currentNode.right != null )
						nextLevelStarter = currentNode.right;
				}
				//add to the thisLevelList
				thisLevelList.add(currentNode.val);
				if( currentNode.left != null )
					q.add(currentNode.left);
				if( currentNode.right != null )
					q.add(currentNode.right);
			}
			answerList.add(thisLevelList);
		}
		//reverse the result list
		List<List<Integer>> reverseAnswerList = new ArrayList<List<Integer>>();
		for( int i = answerList.size()-1; i >= 0 ; i-- ){
			reverseAnswerList.add(answerList.get(i));
		}
		return reverseAnswerList;
    }
}