235. Lowest Common Ancestor of a Binary Search Tree

Description

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).”

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.

Approach

Links

• Binarysearch

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : O(N), where N is the number of nodes in the BST. 
 *    In the worst case we might be visiting all the nodes of the BST.
 * Space complexity : O(N). This is because the maximum amount of space 
 *    utilized by the recursion stack would be N since the height of a 
 *    skewed BST could be N.
 */

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null) {
            return null;
        }
        if(root.val > Math.max(p.val, q.val)) {
            return lowestCommonAncestor(root.left, p, q);
        } else if(root.val < Math.min(p.val, q.val)) {
            return lowestCommonAncestor(root.right, p, q);
        }
        return root;
    }
}

Solution 2

/**
 * Time complexity : O(N), where N is the number of nodes in the BST. 
 *    In the worst case we might be visiting all the nodes of the BST.
 * Space complexity : O(1).
 */

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {

        // Value of p
        int pVal = p.val;

        // Value of q;
        int qVal = q.val;

        // Start from the root node of the tree
        TreeNode node = root;

        // Traverse the tree
        while (node != null) {

            // Value of ancestor/parent node.
            int parentVal = node.val;

            if (pVal > parentVal && qVal > parentVal) {
                // If both p and q are greater than parent
                node = node.right;
            } else if (pVal < parentVal && qVal < parentVal) {
                // If both p and q are lesser than parent
                node = node.left;
            } else {
                // We have found the split point, i.e. the LCA node.
                return node;
            }
        }
        return null;
    }
}

Follow up