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; } }
来源:https://www.cnblogs.com/xiagnming/p/9639223.html