235. 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]

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.

M1: recursive

从root开始找，如果p, q都在左，继续往左找；如果p, q都在右，继续往右找；如果p,q一左一右，说明有符合条件的common ancestor，返回root

time: O(n)　　-- worst case, visiting all nodes, space: O(n)　　-- recursion stack, also worst case

/**
 * 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(p.val > root.val && q.val > root.val) {
            return lowestCommonAncestor(root.right, p, q);
        }
        else if(p.val < root.val && q.val < root.val) {
            return lowestCommonAncestor(root.left, p, q);
        }
        else {
            return root;
        }
    }
}

M2: iterative

time: O(n), space: O(1)

/**
 * 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) {
        TreeNode node = root;
        while(node != null) {
            if(p.val > node.val && q.val > node.val) {
                node = node.right;
            }
            else if(p.val < node.val && q.val < node.val) {
                node = node.left;
            }
            else {
                return node;
            }
        }
        return null;
    }
}