Let me walk you through the different variations of the LCA problem and how to solve each one with slight modifications to the core algorithm.

1. Basic LCA Problem

Problem: Find the LCA of two nodes p and q in a binary tree (both nodes guaranteed to exist).

Solution Approach:

• Recursively traverse the tree
• If we find either p or q, return that node
• If a node has both left and right subtrees returning non-null values, that's the LCA
• Otherwise propagate whichever subtree has a found node

var lowestCommonAncestor = function(root, p, q) {
    function lca(node, p, q) {
    if (!node) return null;
    if (node === p || node === q) return node;
        
    let left = lca(node.left, p, q);
    let right = lca(node.right, p, q);
        
    if (left && right) return node; // Current node is LCA
    return left || right; 
// Propagate whichever side has a found node
    }
    return lca(root, p, q);
};

2. LCA II - Nodes May Not Exist (Leetcode 1644)

Problem: Find LCA of p and q, but return null if either doesn't exist.

Solution Approach:

• First verify both nodes exist in the tree
• Then use the standard LCA algorithm

var lowestCommonAncestor = function(root, p, q) {
   

// Helper to check if node exists in tree

function nodeExists(node, target) {
 if (!node) return false;
 if (node === target) return true;
return nodeExists(node.left, target) || nodeExists(node.right, target);
}
    

// First check if both nodes exist
if (!nodeExists(root, p) || !nodeExists(root, q)) return null;

    // Standard LCA algorithm
    function lca(node) {
        if (!node || node === p || node === q) return node;
        let left = lca(node.left);
        let right = lca(node.right);
        if (left && right) return node;
        return left || right;
    }
    
    return lca(root);
};

3. LCA IV - Multiple Nodes (Leetcode 1676)

Problem: Find LCA of an array of nodes (all guaranteed to exist).

Solution Approach:

• Similar to basic LCA but generalize for multiple nodes
• Use a Set for O(1) lookups of target nodes
• A node is LCA if it has target nodes in both subtrees or is itself a target

var lowestCommonAncestor = function(root, nodes) {
    const targets = new Set(nodes.map(n => n.val));
    
    function lca(node) {
        if (!node) return null;
        if (targets.has(node.val)) return node;
        
        let left = lca(node.left);
        let right = lca(node.right);
        
        if (left && right) return node;
        return left || right;
    }
    
    return lca(root);
};

4. LCA III - With Parent Pointers (Leetcode 1650)

Problem: Find LCA when nodes have parent pointers (no root given).

Solution Approach:

• Find the depth of both nodes from root
• Move the deeper node up to match the other's depth
• Then move both up until they meet

 (using hash set):

var lowestCommonAncestor = function(p, q) {
    const visited = new Set();
    
    // Traverse up from p
    while (p) {
        visited.add(p);
        p = p.parent;
    }
    
    // Traverse up from q until we find a visited node
    while (q) {
        if (visited.has(q)) return q;
        q = q.parent;
    }
    
    return null;
};

Key Insights:

1. Core LCA Algorithm: The recursive approach works beautifully for standard binary trees by checking left and right subtrees.
2. Node Existence Check: For problems where nodes might not exist, we need to first verify their presence.
3. Multiple Nodes: Generalizing to multiple nodes just requires tracking a set of target nodes.
4. Parent Pointers: When we have parent pointers, we can use depth calculation or path tracking to find the intersection.

The beauty is that all these variations build on the same fundamental concept - finding the deepest node that contains all target nodes in its descendants.