You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
{{ message }}
Repository navigation
Expand file tree
/
235. Lowest Common Ancestor of a Binary Search Tree.js
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).”
Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
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.
Note:
All of the nodes' values will be unique.
p and q are different and both values will exist in the BST.
*/
/**
* Leetcode fundamental: 11/5 Update
* Failure:
* 1. Fail to think of condition to find next child node by comparing p.val, q.val and curr.val
*
* Note: Hard to think of stack and while loop solution
* Memoize the solution!
*/
/**
* 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}
*/
varlowestCommonAncestor=function(root,p,q){
// Condition to find left child node (p.val < currNode.val && q.val < currNode.val)
if(p.val<root.val&&q.val<root.val)
returnlowestCommonAncestor(root.left,p,q);
// Codition to find right child node (p.val > currNode.val && q.val > currNode.val)