Leetcode 235. Lowest Common Ancestor of a Binary Search Tree

最新推荐文章于 2024-02-06 06:38:28 发布

转载最新推荐文章于 2024-02-06 06:38:28 发布 · 68 阅读

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原文链接：http://www.cnblogs.com/xiagnming/p/9639223.html

本文介绍如何在二叉搜索树中寻找两节点的最低公共祖先。通过递归方式，利用二叉搜索树特性，当目标节点分别位于当前节点两侧时，则当前节点即为所求。

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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;
    }
}

转载于:https://www.cnblogs.com/xiagnming/p/9639223.html