博客
关于我
剑指offer--树--树中两个结点的最低公共祖先
阅读量:135 次
发布时间:2019-02-26

本文共 2510 字,大约阅读时间需要 8 分钟。

???????????????????????????????

  • ??????????????????????????????????????DFS???????
  • ??????????????????????????????????????????????
  • ??????????????

    import java.util.ArrayList;import java.util.Stack;public class Test2 {    public static void main(String[] args) {        // ????        TreeNode root = new TreeNode(6);        TreeNode left = new TreeNode(2);        left.left = new TreeNode(0);        left.right = new TreeNode(4);        left.right.left = new TreeNode(7);        left.right.right = new TreeNode(9);        root.left = left;        TreeNode right = new TreeNode(8);        right.left = new TreeNode(3);        right.left.right = new TreeNode(5);        root.right = right;        System.out.println(lowestCommonAncestorII(root, left, right));    }    public static TreeNode lowestCommonAncestorII(TreeNode root, TreeNode p, TreeNode q) {        List
    pPath = findPath(root, p); List
    qPath = findPath(root, q); // ??????????null if (pPath == null || qPath == null) { return null; } // ??????????????? int minLength = Math.min(pPath.size(), qPath.size()); TreeNode common = null; for (int i = 0; i < minLength; i++) { if (pPath.get(i) == qPath.get(i)) { common = pPath.get(i); } else { break; } } return common; } private static List
    findPath(TreeNode root, TreeNode target) { List
    path = new ArrayList<>(); Stack
    stack = new Stack<>(); stack.push(root); while (!stack.isEmpty()) { TreeNode node = stack.pop(); if (node == target) { // ????????????path? while (!stack.isEmpty()) { path.add(stack.pop()); } return path; } if (node.left != null) { stack.push(node.left); } if (node.right != null) { stack.push(node.right); } } return null; }}

    ????

  • findPath ???

    • ?????????????DFS?????????????????
    • ???????????????????????????????????
  • lowestCommonAncestorII ???

    • ?? findPath ??????????p?q????
    • ?????????????????????
    • ?????????????????????????????????????
  • ??????

    • main ???????????? lowestCommonAncestorII ???????????
    • lowestCommonAncestorII ?????????????????????????????
    • findPath ????????DFS?????????

    ????????????O(n)???????????????????????????????

    转载地址:http://tnzu.baihongyu.com/

    你可能感兴趣的文章
    Openlayers高级交互(4/20):手绘多边形,导出KML文件,可以自定义name和style
    查看>>
    Openlayers高级交互(5/20):右键点击,获取该点下多个图层的feature信息
    查看>>
    Openlayers高级交互(6/20):绘制某点,判断它是否在一个电子围栏内
    查看>>
    Openlayers高级交互(7/20):点击某点弹出窗口,自动播放视频
    查看>>
    Openlayers高级交互(8/20):选取feature,平移feature
    查看>>
    Openlayers:DMS-DD坐标形式互相转换
    查看>>
    openlayers:圆孔相机根据卫星经度、纬度、高度、半径比例推算绘制地面的拍摄的区域
    查看>>
    OpenLDAP(2.4.3x)服务器搭建及配置说明
    查看>>
    OpenLDAP编译安装及配置
    查看>>
    Openmax IL (二)Android多媒体编解码Component
    查看>>
    OpenMCU(一):STM32F407 FreeRTOS移植
    查看>>
    OpenMCU(三):STM32F103 FreeRTOS移植
    查看>>
    OpenMCU(三):STM32F103 FreeRTOS移植
    查看>>
    OpenMCU(二):GD32E23xx FreeRTOS移植
    查看>>
    OpenMCU(五):STM32F103时钟树初始化分析
    查看>>
    OpenMCU(四):STM32F103启动汇编代码分析
    查看>>
    OpenMetadata 命令执行漏洞复现(CVE-2024-28255)
    查看>>
    OpenMMLab | AI玩家已上线!和InternLM解锁“谁是卧底”新玩法
    查看>>
    OpenMMLab | S4模型详解:应对长序列建模的有效方法
    查看>>
    OpenMMLab | 【全网首发】Llama 3 微调项目实践与教程(XTuner 版)
    查看>>