LeetCode 之 Binary Tree Level Order Traversal II
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题目链接:Binary Tree Level Order Traversal II
这虽然是道”easy”的题,做的时候整个思路有不是很复杂,但是编写代码的时候,主要是对list还不够熟练,出现了很多问题,导致做了很久才做出来。
思路:分别得出左右子树对应的list,然后合并成整个树的list,得出的list的顺序需要反转。其中利用到树的深度。
代码如下:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * }*/class Solution { //对list进行反转 public List<List<Integer>> levelOrderBottom(TreeNode root) { List<List<Integer>> first = new ArrayList<>(); first = levelOrderBottomreverse(root); Collections.reverse(first); return first; } public List<List<Integer>> levelOrderBottomreverse(TreeNode root) { int depth = maxDepth(root); List<List<Integer>> re = new ArrayList<>(); if(root==null) return re; int depthl = maxDepth(root.left); int depthr = maxDepth(root.right); ArrayList<Integer> sub = new ArrayList<Integer>(); sub.add(root.val); re.add(sub); List<List<Integer>> rel = levelOrderBottomreverse(root.left); List<List<Integer>> rer = levelOrderBottomreverse(root.right); for(int i=1;i<depth;i++){ sub = new ArrayList<Integer>(); if(i<=depthl) sub.addAll(rel.get(i-1)); if(i<=depthr) sub.addAll(rer.get(i-1)); re.add(sub); } return re; } //计算树的深度 public int maxDepth(TreeNode root) { int maxdepth = 0; if(root==null) return 0; int maxl = maxDepth(root.left); int maxr = maxDepth(root.right); maxdepth = 1 + ((maxl>maxr)? maxl:maxr); return maxdepth; }}
其中初始化定义List<List<Integer>>
时,需要注意
初始化过程是:
List<List<Integer>> list = new ArrayList<List<Integer>>();//或者Java 1.7以后List<List<Integer>> list = new ArrayList<>();
同时,查看了其他人的解题思路:(今天有点累了,明天补上)
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