235. Lowest Common Ancestor of a Binary Search Tree

Difficulty: Easy

Related Topics: Tree

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

According to the : “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).”

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.

Example 3:

Input: root = [2,1], p = 2, q = 1
Output: 2

Constraints:

• The number of nodes in the tree is in the range [2, 10].
• -10 <= Node.val <= 10
• All Node.val are unique.
• p != q
• p and q will exist in the BST.

Solution

Language: Java

class Solution {

    // 最大公共祖先定义：是两个节点所有公共祖先中离根节点最远的
    // BST 的性质：左子树节点 < 根节点
    //            右子树节点 > 根节点
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // LAC 只能在左子树中
        if (p.val < root.val && q.val < root.val) {
            return lowestCommonAncestor(root.left, p, q);
        }

        // LAC 只能在右子树中
        if (p.val > root.val && q.val > root.val) {
            return lowestCommonAncestor(root.right, p, q);
        }

        // p 和 q 一个在左子树中，一个在右子树中
        // 那么 root 就是 LCA
        return root;
    }
}