235. Lowest Common Ancestor of a Binary Search Tree

Hint - Recursion

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
    {
        // If both p and q are less than root's value, then LCA must be in the left subtree

        // If both p and q are greater than root's value, then LCA must be in the right subtree

        // If p and q are on different sides of root, or one of them is the root, then root is the LCA
    }
};

• Recursion

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
    {
        if (p->val < root->val && q->val < root->val)
        {
            return lowestCommonAncestor(root->left, p, q);
        }

        if (p->val > root->val && q->val > root->val)
        {
            return lowestCommonAncestor(root->right, p, q);
        }
        return root;
    }
};

• T: $O(N)$
• S: $O(N)$

Hint - Iteration

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
    {
        // Start with the root node

        // Loop until we find the lowest common ancestor
        while ()
        {
            // Get the value of the current node

            // If both p and q are greater than parentVal, LCA must be in the right subtree
            if ()
            {
            }
            // If both p and q are less than parentVal, LCA must be in the left subtree
            else if ()
            {
            }
            // If p and q are on different sides of the current node, or one of them is the current node,
            // then the current node is the lowest common ancestor
            else
            {
            }
        }
        // This line is never reached because the loop always returns a node
    }
};

• Iteration

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
    {
        TreeNode* node = root;
        while (true)
        {
            int parentVal = node->val;
            if (parentVal < p->val && parentVal < q->val)
            {
                node = node->right;
            }
            else if (parentVal > p->val && parentVal > q->val)
            {
                node = node->left;
            }
            else
            {
                return node;
            }
        }
        return node;
    }
};

• T: $O(N)$
• S: $O(1)$