二叉树最长距离
tag: 二叉树
思路:
最长距离一定是两个叶子节点之间的距离
=> 两个叶子节点必定以某个节点为根节点
=> 因此用DFS思路自底向上计算经过每一个节点的最长距离,取其最大值
以每个节点为根节点的最长距离 = 左子树的高度+右子树的高度
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 | package com.zhaochao.tree; /** * Created by zhaochao on 17/1/25. */ public class MaxDistanceOfBT { int max = 0 ; public int maxDistance(TreeNode root) { if (root == null ) { return 0 ; } if (root.left == null && root.right == null ) { return 0 ; } //双枝 max = Math.max(getHeight(root.left) + getHeight(root.right),max); //返回单枝 return Math.max(getHeight(root.left),getHeight(root.right)); } public int getHeight(TreeNode root) { if (root == null ) { return 0 ; } int left = getHeight(root.left); int right = getHeight(root.right); return Math.max(left, right) + 1 ; } public static void main(String[] args) { TreeNode root = new TreeNode( 0 ); TreeNode node1 = new TreeNode( 1 ); TreeNode node2 = new TreeNode( 2 ); TreeNode node3 = new TreeNode( 3 ); TreeNode node4 = new TreeNode( 3 ); TreeNode node5 = new TreeNode( 3 ); TreeNode node6 = new TreeNode( 3 ); TreeNode node7 = new TreeNode( 3 ); TreeNode node8 = new TreeNode( 3 ); TreeNode node9 = new TreeNode( 3 ); root.left = node1; root.right = node2; node1.left = node3; node3.left = node8; node8.left = node9; node1.right = node5; node5.right = node6; node2.right = node4; MaxDistanceOfBT test = new MaxDistanceOfBT(); test.maxDistance(root); System.out.println(test.max); } } |
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】博客园社区专享云产品让利特惠,阿里云新客6.5折上折
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步