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 thedefinition 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 allowa 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

Note:

• All of the nodes' values will be unique.

• p and q are different and both values will exist in the BST.

Solution

If root value is bigger than p node and bigger than value of q node, we go to left. If value is smaller, we go to right. Otherwise we return the node.

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root.val < p.val && root.val < q.val){
            return lowestCommonAncestor(root.right, p, q);
        } else if(root.val > p.val && root.val > q.val){
            return lowestCommonAncestor(root.left, p, q);
        } else{
            return root;
        }
    }
}

Another solution using min and max functions.

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

        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);
        } else {
            return root;
        }

    }
}

Non recursive solution.

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        while (root.val > p.val && root.val > q.val) root = root.left;
        while (root.val < p.val && root.val < q.val) root = root.right;
        return root;
    }
}