• Leetcode 235. Lowest Common Ancestor of a Binary Search Tree


    题目链接

    https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/

    题目描述

    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]

            _______6______
           /              
        ___2__          ___8__
       /              /      
       0      _4       7       9
             /  
             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.

    题解

    因为是搜索二叉树,如果两个节点的值一个比root大,一个比root小,就说明根节点是它们的公共结点,反之,它们必然位于同一侧,递归遍历即可。

    代码

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
            if (root == null) {
                return null;
            }
            if (root.val > p.val && root.val > q.val) {
                return lowestCommonAncestor(root.left, p, q);
            } else if (root.val < p.val && root.val < q.val) {
                return lowestCommonAncestor(root.right, p, q);
            }
            return root;
        }
    }
    
    
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  • 原文地址:https://www.cnblogs.com/xiagnming/p/9639223.html
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