# Provide a more general way to handle such problem, not limit to BST

• While at begining i didn't notice it is a BST, which can be evaluate by its value. So i came up with this solution without using node's value.

``````public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) return null;
if ((p == root && isChild(root, q)) ||
(q == root && isChild(root, p)) ||
(isChild(root.left, p) && isChild(root.right, q)) ||
(isChild(root.right, p) && isChild(root.left, q))) {
return root;
}
TreeNode res = null;
if((res = lowestCommonAncestor(root.left, p, q)) != null) return res;
if((res = lowestCommonAncestor(root.right, p, q)) != null) return res;
return null;
}

public boolean isChild(TreeNode root, TreeNode child) {
if (root == child) return true;
if (root == null) return false;
return isChild(root.left, child) || isChild(root.right, child);
}
}``````

• This solution runs in O(n^2) doesn't it?

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