How to solve Leetcode 235. Lowest Common Ancestor of a Binary Search Tree

How to solve Leetcode 235. Lowest Common Ancestor of a Binary Search Tree

Finding the lowest common ancestor node of two given nodes in a binary search tree

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Nhut Nguyen
Aug 29, 2022·

2 min read

Problem statement

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

binarysearchtree_improved.png

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

binarysearchtree_improved.png

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: Recursion

Note that in a BST, the values of a node and its children left and right satisfy

left.value < node.value < right.value.

It lets you know which branch (left or right) of the root the nodes p and q belong to.

Code

#include <iostream>
using namespace std;
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if (p->val < root->val && q->val < root->val) {
        return lowestCommonAncestor(root->left, p, q);
    } else if (root->val < p->val && root->val < q->val) {
        return lowestCommonAncestor(root->right, p, q);
    }
    return root;
}

int main() {
    TreeNode zero(0);
    TreeNode three(3);
    TreeNode five(5);
    TreeNode four(4);
    four.left = &three;
    four.right = &five;
    TreeNode two(2);
    two.left = &zero;
    two.right = &four;
    TreeNode seven(7);
    TreeNode nine(9);
    TreeNode eight(8);
    eight.left = &seven;
    eight.right = &nine;
    TreeNode six(6);
    six.left = &two;
    six.right = &eight;

    cout << lowestCommonAncestor(&six, &two, &eight)->val << endl;
    cout << lowestCommonAncestor(&six, &two, &four)->val << endl;
    cout << lowestCommonAncestor(&two, &two, &zero)->val << endl;
}

Output:
6
2
2

Complexity

  • Runtime: O(logN) (the height of the tree), where N is the number of nodes.

  • Extra space: O(1).

References