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).”

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.

Solution & Analysis

Using attribution of a BST

/**
 * 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) {

        // Value of current node or parent node.
        int parentVal = root.val;

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

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

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

Iterative

Same code as Lowest Common Ancestor of a Binary Tree

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null || root == p || root == q) return root;
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if(left != null && right != null) return root;
        return left != null ? left : right;
    }
}

Reference

https://www.lintcode.com/problem/lowest-common-ancestor-ii/description

https://www.lintcode.com/problem/lowest-common-ancestor-iii/description