zoukankan      html  css  js  c++  java
  • 235. Lowest Common Ancestor of a Binary Search Tree

    Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

    According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

    Given binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]

    Example 1:

    Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
    Output: 6
    Explanation: The LCA of nodes 2 and 8 is 6.
    

    Example 2:

    Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
    Output: 2
    Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
    

    Note:

    • All of the nodes' values will be unique.
    • p and q are different and both values will exist in the BST.
      /**
       * Definition for a binary tree node.
       * public class TreeNode {
       *     int val;
       *     TreeNode left;
       *     TreeNode right;
       *     TreeNode(int x) { val = x; }
       * }
       */
      public class Solution {
          public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
              if(root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left, p, q);
              if(root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q);
              return root;
          }
      }

      这个解法简直钛蚌

    • 先看recursive。从根节点出发,分别去搜索两个节点,但这里搜索不用瞎搜索,需要利用BST的性质,比较p->val,q->val和root->val的关系,决定往左边搜还是右边搜。如果root->val>p->val同时root->val>q->val,说明p,q都在root的左侧,因此把root=root->left;如果root->val< p->val同时root->val < q->val,说明p,q都在root的右侧,因此把root=root->right;如果root->val>p->val同时root->val< q->val,或root->val< p->val同时root->val> q->val说明p,q在root的两侧,说明root就是LCA.
      ————————————————
      版权声明:本文为CSDN博主「xinqrs01」的原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接及本声明。
      原文链接:https://blog.csdn.net/xinqrs01/article/details/54999271

  • 相关阅读:
    APIO2007 动物园
    SCOI2010 股票交易
    USACO13NOV No Change G
    洛谷 P3694 邦邦的大合唱站队
    洛谷 P6196 3月月赛 ERR1 代价
    洛谷月赛 ERR1 代价
    Splay 学习笔记
    HNOI2009 梦幻布丁
    乘法逆元
    【洛谷】【二分答案+最短路】P1462 通往奥格瑞玛的道路
  • 原文地址:https://www.cnblogs.com/wentiliangkaihua/p/11444491.html
Copyright © 2011-2022 走看看