LeetCode 235. Lowest Common Ancestor of a Binary Search Tree

Question

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).”

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.

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.

Source: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

Solution

// If BST is balanced, Time Complexity O(log(n))
// Worst Time Complexity O(n)
// return the lowest common ancestor
// with an assumption that p and q must exist in tree
private TreeNode dfs(TreeNode root, TreeNode p, TreeNode q) {
    if (p.val < root.val && q.val < root.val) { // both p, q in the left tree
        return dfs(root.left, p, q);
    } else if (p.val > root.val && q.val > root.val) { // both p, q in the right tree
        return dfs(root.right, p, q);
    } else {
        // p or q is current root,
        // or p and q are in two different subtrees respectively
        // in either case, current root is the lowest common ancestor
        return root;
    }
}

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    return dfs(root, p, q);
}

private TreeNode lca = null;

// an accelerated brutal force DFS, so has higher time complexity
// but works for cases that p and q may not exist in the tree
private int dfs2(TreeNode root, TreeNode p, TreeNode q) {
    if (root == null) {
        return 0;
    }

    int count = 0;
    if (root == p || root == q) {
        count++;
    }

    if (p.val < root.val && q.val < root.val) {
        count += dfs2(root.left, p, q);
    } else if (p.val > root.val && q.val > root.val) {
        count += dfs2(root.right, p, q);
    } else {
        count += dfs2(root.left, p, q);
        count += dfs2(root.right, p, q);
    }

    if (count == 2 && lca == null) {
        lca = root;
    }
    return count;
}

public TreeNode lowestCommonAncestor2(TreeNode root, TreeNode p, TreeNode q) {
    lca = null;
    dfs2(root, p, q);
    return lca;
}