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
    Objective-C实现鸡兔同笼问题(附完整源码)
    查看>>
    Objective-c正确的写法单身
    查看>>
    Objective-C语法之代码块(block)的使用
    查看>>
    ObjectMapper - 实现复杂类型对象反序列化(天坑!)
    查看>>
    ObjectProperty 类的使用
    查看>>
    Object常用方法
    查看>>
    Object方法的finalize方法
    查看>>
    Object类有哪些方法,hashcode方法的作用,为什么要重写hashcode方法?
    查看>>
    Objenesis创建类的实例
    查看>>
    OBObjective-c 多线程(锁机制) 解决资源抢夺问题
    查看>>
    OBS studio最新版配置鉴权推流
    查看>>
    Obsidian的使用-ChatGPT4o作答
    查看>>
    Obsidian笔记记录GPT回复的数学公式无缝转化插件Katex to mathjax
    查看>>
    ObsoleteAttribute 可适用于除程序集、模块、参数或返回值以外的所有程序元素。 将元素标记为过时可以通知用户:该元素在产品的未来版本中将被移除。...
    查看>>
    OC Xcode快捷键
    查看>>
    oc 中的.m和.mm文件区别
    查看>>
    OC 中的重写 OC中没有重载 以及隐藏
    查看>>
    OC 内存管理黄金法则
    查看>>
    oc57--Category 分类
    查看>>
    occi库在oracle官网的下载针对vs2008
    查看>>