leetcode235 - Lowest Common Ancestor of a Binary Search Tree - easy

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.

利用BST特性。
1.如果root值在要找的两个数中间，说明两个节点分布在左右子树里，当前的根就是LCA。
2.如果root值和两个数其中一个相等，说明两个数的节点是亲子关系，当前的根就是LCA。
以上面两条规则为递归出口，然后做递归。
类似问题：Lowest Common Ancestor of a Binar Tree: https://www.cnblogs.com/jasminemzy/p/9698483.html

实现：

/**
 * 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 || root.val == p.val || root.val == q.val) {
            return root;
        }
        if (root.val > p.val && root.val < q.val || root.val < p.val && root.val > q.val) {
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        return left != null ? left : right;
    }
}

原文地址：https://www.cnblogs.com/jasminemzy/p/9736893.html