[LeetCode] 235. Lowest Common Ancestor of a Binary Search Tree

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).”

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, 105].
-109 <= Node.val <= 109
All Node.val are unique.
p != q
p and q will exist in the BST.

二叉搜索树的最小公共祖先。

影子题236。题意是给一个二叉搜索树（BST）和两个节点p和q，请找出两个节点的最小公共祖先。思路是递归，因为BST的性质，所以任何节点的左孩子一定比当前节点小，右孩子一定比当前节点大。明白这一点，代码就不难写。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
 3         if (root.val > p.val && root.val > q.val) {
 4             return lowestCommonAncestor(root.left, p, q);
 5         } else if (root.val < p.val && root.val < q.val) {
 6             return lowestCommonAncestor(root.right, p, q);
 7         } else {
 8             return root;
 9         }
10     }
11 }

JavaScript实现

 1 /**
 2  * @param {TreeNode} root
 3  * @param {TreeNode} p
 4  * @param {TreeNode} q
 5  * @return {TreeNode}
 6  */
 7 var lowestCommonAncestor = function (root, p, q) {
 8     if (root.val > p.val && root.val > q.val) {
 9         return lowestCommonAncestor(root.left, p, q);
10     } else if (root.val < p.val && root.val < q.val) {
11         return lowestCommonAncestor(root.right, p, q);
12     } else {
13         return root;
14     }
15 };