235&236. Lowest Common Ancestor of a Binary (Search) Tree

236. Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

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 = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

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 tree.

My Solutions:

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    if (root == null || root == p || root == q) return root;
    
    TreeNode left = lowestCommonAncestor(root.left, p, q);
    TreeNode right = lowestCommonAncestor(root.right, p, q);
    
    // it means the p and q has been found on the root's left and right
    if (left != null && right != null) return root;
    if (left != null) return left;
    if (right != null) return right;
    return null;
}

235. Lowest Common Ancestor of a Binary Search Tree

235和236的区别是235是BST，因此可以只在root的一侧找p和。因为如果p和q的值在root两边，说明root就是结果。

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    if (root == null || root == p || root == q) return root;
    if (root.val > p.val && root.val > q.val) {
        return lowestCommonAncestor(root.left, p, q);
    } else if (root.val < p.val && root.val < q.val) {
        return lowestCommonAncestor(root.right, p, q);
    } else {
        return root;
    }
}