235. Lowest Common Ancestor of a Binary Search Tree (Easy)

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

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]

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.

Solutions

1.

class Solution {

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return null;
        }

        List<TreeNode> pTrack = new ArrayList<>();
        List<TreeNode> qTrack = new ArrayList<>();

        // Get the track to p and q nodes
        searchNode(root, p, pTrack);
        searchNode(root, q, qTrack);

        int lenp = pTrack.size();
        int lenq = qTrack.size();

        int len = Math.min(lenp, lenq);

        // Find the first different node before which is the lowest common ancestor.
        for (int i = 0; i < len; i++) {
            if (pTrack.get(i).val != qTrack.get(i).val) {
                return pTrack.get(i - 1);
            }
        }

        // Certain one is the prefix of another.
        return pTrack.get(len - 1);
    }

    private void searchNode(TreeNode root, TreeNode x, List<TreeNode> track) {
        track.add(root);

        if (root == null) {
            return;
        }

        if (root.val == x.val) {
            return;
        }

        if (root.val > x.val) {
            searchNode(root.left, x, track);
        } else {
            searchNode(root.right, x, track);
        }
    }
}

2.

class Solution {

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // Value of current node or parent node.
        int parentVal = root.val;

        // Value of p
        int pVal = p.val;

        // Value of q;
        int qVal = q.val;

        // If both p and q are greater than parent
        if (pVal > parentVal && qVal > parentVal) {
            return lowestCommonAncestor(root.right, p, q);
        }

        // If both p and q are lesser than parent
        if (pVal < parentVal && qVal < parentVal) {
            return lowestCommonAncestor(root.left, p, q);
        }

        // We have found the split point, i.e. the LCA node.
        return root;
    }
}

Incorrect Solutions

References