Problem
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 Tree
that has both p
and q
as descendants (where we allow a node to be a descendant of itself).”
Input
root of binary search tree, node p and node q
Output
Lowest common ancestor of nodes p and q.
Constraints
- 2 <= Number of Nodes in the Tree ≤ 100000
- -10^9 <= Value of Nodes ≤ 10^9
- All nodes are unique
Sample Input
root = [6, 2, 8, 0, null, null, 10, null, 1, 9, 22] p = 8, q = 22
Sample Output
8
Code Implementation
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (root == NULL) {
return NULL;
}
if (root == p || root == q) {
return root;
}
TreeNode *lNode = lowestCommonAncestor(root->left, p, q);
TreeNode *rNode = lowestCommonAncestor(root->right, p, q);
if (lNode && rNode) {
return root;
}
return lNode != NULL ? lNode : rNode;
}
};
Time complexity: O(n), where n is the number of nodes in tree. This is because it’s a simple tree traversal.