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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

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        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Iterative solution

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || p == null || q == null){return null;}
        while ((root.val - p.val) * (root.val - q.val) > 0){
            root = p.val < root.val ? root.left : root.right;
        }
        return root;
    }
}

Recursive solution

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public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || p == null || q == null){return null;}
        if ((root.val - p.val) * (root.val - q.val) <= 0){
            return root;
        }else{
            if (p.val < root.val){
                return lowestCommonAncestor(root.left, p,q);
            }else{
                return lowestCommonAncestor(root.right, p,q);
            }
        }
    }
}

Same one line solution

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public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || p == null || q == null){return null;}
        return ((root.val - p.val) * (root.val - q.val) <= 0) ? root :
        lowestCommonAncestor(p.val < root.val ? root.left : root.right, p, q);
    }
}