235. 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 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).”

        _______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.

Solution:

Method 1: recursive

This is a BST. Given any node, the children in its left subtree are always less than it. The children in its right subtree are always larger than it.

If both p and q are smaller than root, the LCA is in its left subtree.

If both p and q are larger than root, the LCA is in its right subtree.

Otherwise, root is the LCA.



Code:

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

Method 2: non-recersive

The same idea with method 1.

We use a while loop to check whether both p and q are greater or less than root. And decide where to go and do further search.

If p and q are not in the same side, root is LCA.

Time complexity is O(h) where h is the height of the tree.



Code:

/**
 * 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) {
        while ((root.val - p.val) * (root.val - q.val) > 0) {
            if (root.val > p.val) {
                root = root.left;
            }
            else {
                root = root.right;
            }
        }
        return root;
    }
}