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Lowest Common Ancestor of a Binary Search Tree⚓︎

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Description⚓︎

Given a binary search tree (BST), find the lowest common ancestor (LCA) node 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:

Ex1

  • 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:

Ex2

  • 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

Constraints:

  • The number of nodes in the tree is in the range [2, 10^5].
  • -10^9 <= Node.val <= 10^9
  • All Node.val are unique.
  • p != q
  • p and q will exist in the BST.

Solution⚓︎

Recursive Solution⚓︎

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (p->val < root->val && q->val < root->val) {
            return lowestCommonAncestor(root->left, p, q);
        }
        if (p->val > root->val && q->val > root->val) {
            return lowestCommonAncestor(root->right, p, q);
        }
        return root;
    }
};
  • Worst time complexity: \(O(n)\);
  • Worst space complexity: \(O(n)\).

Iterative Solution⚓︎

root go from top to bottom:

  • If p is encountered first, then p is the answer.
  • If q is encountered first, then q is the answer.
  • If root is between the values of p and q, it doesn't matter who is bigger or smaller, as long as root is in the middle, then root is the answer.
  • If root is to the left (smaller) of the values of p~q, then root moves to the right.
  • If root is to the right (bigger) of the values of p~q, then root moves to the left.
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        while (root->val != p->val && root->val != q->val) {
            if (min(p->val, q->val) < root->val && root->val < max(p->val, q->val))
                break;
            root = root->val < min(p->val, q->val) ? root->right : root->left;
        }
        return root;
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(1)\).