Lowest Common Ancestor of a Binary Search Tree LeetCode Solution

Minimum Cost to Merge Stones

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 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.

Example 3:

Input: root = [2,1], p = 2, q = 1
Output: 2


  • 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.

Lowest Common Ancestor of a Binary Search Tree Solutions

Time: O(h)
Space: O(h)


class Solution {
  TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if (root->val > max(p->val, q->val))
      return lowestCommonAncestor(root->left, p, q);
    if (root->val < min(p->val, q->val))
      return lowestCommonAncestor(root->right, p, q);
    return root;


 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);
    if (root.val < Math.min(p.val, q.val))
      return lowestCommonAncestor(root.right, p, q);
    return root;



class Solution:
  def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
    if root.val > max(p.val, q.val):
      return self.lowestCommonAncestor(root.left, p, q)
    if root.val < min(p.val, q.val):
      return self.lowestCommonAncestor(root.right, p, q)
    return root

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