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

  • 0

    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); 

  • 1

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

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