235. Lowest Common Ancestor of a Binary Search 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:
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:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
// base case
if (root == null){
return null;
}
if (root == p || root == q){
return root;
}
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if (left != null && right != null){
return root;
}
if (left == null && right == null){
return null;
}
if (left != null){
return left;
}else{
return right;
}
}
}
/*
case 1 : root.left == null && root.right == null -> return null
case 2 : root.left != null || root.right != null -> return not null side like root.left or root.right
case 3 : root.left && root.right != null return root;
*/
// TC: O(n)
// SC: O(n)
Binary Search Tree (BST): Due to the inherent property of a BST (left child is smaller, and right child is larger), finding the LCA is relatively easier and can be done in a more straightforward manner.
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