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
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
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
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
andq
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), whereN
is the number of nodes.Extra space:
O(1)
.
References
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