107. Binary Tree Level Order Traversal II#1(Done)

    xiaoxiao2021-12-14  18

    Solution

    /** * 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>> result = new LinkedList<List<Integer>>(); if (root == null) { return result; } Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root); while (!queue.isEmpty()) { int n = queue.size(); List<Integer> list = new LinkedList<Integer>(); while (n-- > 0) { TreeNode tmp = queue.poll(); list.add(tmp.val); if (tmp.left != null) { queue.offer(tmp.left); } if (tmp.right != null) { queue.offer(tmp.right); } } result.add(0, list); } return result; } }

    Problem#1 * list中的add可以在指定位置加数据 * 用深度优先搜索和广度优先搜索两种方法实现

    转载请注明原文地址: https://ju.6miu.com/read-970978.html

    最新回复(0)