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,null,null,15,7],
3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3]
]
问题：此题和102一个原理，只是102题输出是从上到下，该题目是从下到上
思想：将ArrayList换成LinkedList，每次添加时调用add(0,levelList)，将结果添加到List的头部
/** * 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) { List<List<Integer>> res=new LinkedList<List<Integer>>(); //ArrayList换LingkedList可以使每次添加的levelList到头部 if(root==null) return res; Queue<TreeNode> q=new LinkedList(); q.add(root); while(!q.isEmpty()){ List<Integer> levelList=new ArrayList();//每层list int size=q.size(); for(int i=0;i<size;i++){ TreeNode node=q.poll(); levelList.add(node.val); if(node.left!=null) q.add(node.left); if(node.right!=null) q.add(node.right); } res.add(0, levelList);//将levelList添加到res的头部，自然形成从下到上的顺序 } return res; } }
