1650. Lowest Common Ancestor of a Binary Tree III

The first solution of this problem can be based on the 236. Lowest Common Ancestor of a Binary Tree too:

    public Node lowestCommonAncestor(Node p, Node q) {
       Node root = p;
        while(root.parent!=null)
            root = root.parent;
        return helper(root, p, q);
    }
    private Node helper(Node root, Node p, Node q){
        if(root==null)
            return null;
        if(root==p || root==q)
            return root;
        Node left = helper(root.left, p, q);
        Node right = helper(root.right, p, q);
        if(left==null)
            return right;
        else if(right==null)
            return left;
        else
            return root;
    }

Because every node of the tree has a pointer to their parent, so this problem can be looked at a linked list too. x and y move forward to root, when meet root, x=q, y=p, at last, when x meet y, the node is their lowest common parent.

    public Node lowestCommonAncestor(Node p, Node q) {
        Node x=p, y=q;
        while(x!=y){
            if(x.parent!=null)
                x=x.parent;
            else
                x = q;
            if(y.parent!=null)
                y=y.parent;
            else
                y=p;
        }
        return x;
    }

 The Recursive solution:

class Solution {
    Node originP, originQ;
    public Node lowestCommonAncestor(Node p, Node q) {
        originP=p;
        originQ = q;
        return helper(p, q);
    }
    
    private Node helper(Node p, Node q){
        if(p==q)
            return p;
        if(p.parent==null)
            p=originQ;
        else 
            p = p.parent;
        if(q.parent==null)
            q=originP;
        else
            q = q.parent;

        return helper(p,q);
    }
}