• LeetCode: Lowest Common Ancestor of a Binary Search Tree 解题报告


    https://leetcode.com/submissions/detail/32662938/

    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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

            _______6______
           /              
        ___2__          ___8__
       /              /      
       0      _4       7       9
             /  
             3   5
    

    For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

    Leetcode又出新题啦

    SOLUTION 1:

    使用递归可以轻松解决此问题。对于此题我们可以分为三种情况讨论:

    1. P, Q都比root小,则LCA在左树,我们继续在左树中寻找LCA

    2. P, Q都比root大,则LCA在右树,我们继续在右树中寻找LCA

    3. 其它情况,表示P,Q在root两边,或者二者其一是root,或者都是root,这些情况表示root就是LCA,直接返回root即可。

    代码如下:

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 public class Solution {
    11     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    12         // 1439.
    13         // 1. Two nodes are on left, go left.
    14         // 2. Two nodes are on right, go right.
    15         // 3. They are in both sides, return root.
    16         // if root == null, return null.
    17         if (root == null) {
    18             return null;
    19         }
    20         
    21         if (p.val < root.val && q.val < root.val) {
    22             return lowestCommonAncestor(root.left, p, q);
    23         } else if (p.val > root.val && q.val > root.val) {
    24             return lowestCommonAncestor(root.right, p, q);
    25         } else {
    26             return root;
    27         }
    28     }
    29 }
    View Code
  • 相关阅读:
    10天学安卓-第八天
    10天学安卓-第七天
    10天学安卓-第六天
    10天学安卓-第五天
    10天学安卓-第四天
    10天学安卓-第三天
    透过 Cucumber 学习 BDD
    应对复杂软件的思考
    管理任务就是管理时间
    Running Dubbo On Spring Boot
  • 原文地址:https://www.cnblogs.com/yuzhangcmu/p/4638649.html
Copyright © 2020-2023  润新知