235. Lowest Common Ancestor of a Binary Search Tree

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235. Lowest Common Ancestor of a Binary Search Tree

Description

Difficulty: Medium

Related Topics: Tree, Depth-First Search, Binary Search Tree, Binary Tree

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:

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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:

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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:

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

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.

Hints/Notes

  • 2023/08/29
  • It’s binary search tree, we can use the property of this data structure
  • 0x3F’s solution(checked)

Solution

Language: C++

Cleaner 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) {
        int x = root->val;
        if (x > p->val && x > q->val) {
            return lowestCommonAncestor(root->left, p, q);
        }
        if (x < p->val && x < q->val) {
            return lowestCommonAncestor(root->right, p, q);
        }
        return root;
    }
};