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235-lowestCommonAncestorOfABinarySearchTree.js
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235-lowestCommonAncestorOfABinarySearchTree.js
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//URL--
// https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
//INSTRUCTIONS--
/* Given a binary search tree (BST), find the lowest common ancestor (LCA) 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, 105].
-109 <= Node.val <= 109
All Node.val are unique.
p != q
p and q will exist in the BST.
*/
//SOLUTION--
/*
If the value of the root node is the value of p, return p
If the value of the root node is the value of q, return q
If the value of the root node is more than the value of p and q,
set the value of the root node to the left node
If the value of the root node is less than the value of p and q,
set the value of the root node to the right node
else, return the current node
*/
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @param {TreeNode} p
* @param {TreeNode} q
* @return {TreeNode}
*/
// const lowestCommonAncestor = function (root, p, q) {
// while (root !== undefined) {
// // If the value of the root node is the value of p, return p
// if (root.val === p.val) {
// return p
// }
// // If the value of the root node is the value of q, return q
// if (root.val === q.val) {
// return q
// }
// // If the value of the root node is more than the value of p and q,
// if (root.val > p.val && root.val > q.val) {
// //set the value of the root node to the left node
// root = root.left
// }
// // If the value of the root node is less than the value of p and q,
// else if (root.val < p.val && root.val < q.val) {
// // set the value of the root node to the right node
// root = root.right
// }
// //if the p or q is less, and the other node is more, return the root
// else {
// return root
// }
// }
// }
/*
Optimizations
I realized that the first two checks (check if the root is equal to p or q) are unnecesary because they also essentailly return the root, so I removed them
*/
/**
* @param {TreeNode} root
* @param {TreeNode} p
* @param {TreeNode} q
* @return {TreeNode}
*/
const lowestCommonAncestor = function (root, p, q) {
while (root !== undefined) {
// If the value of the root node is more than the value of p and q,
if (root.val > p.val && root.val > q.val) {
//set the value of the root node to the left node
root = root.left
}
// If the value of the root node is less than the value of p and q,
else if (root.val < p.val && root.val < q.val) {
// set the value of the root node to the right node
root = root.right
}
//else, return the root
else {
return root
}
}
}
//TESTCASES--
console.log(lowestCommonAncestor([6, 2, 8, 0, 4, 7, 9, null, null, 3, 5], 2, 8), 2);
console.log(lowestCommonAncestor([6, 2, 8, 0, 4, 7, 9, null, null, 3, 5], 2, 4), 6);
console.log(lowestCommonAncestor([2, 1], 2, 1), 2);