Google Interview Question


Country: India
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
2
of 2 vote

just xor the corresponding elements of the 2 trees

- Setu March 22, 2019 | Flag Reply
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0
of 0 vote

Level order/pre-order serialization/deserialization with diff. But, there is a problem, for example, when you have 1 2 3 and 4 5 # you will get diff as 3 3 3. Assume, you get 1 2 3 and 3 3 3 and you have to generate 4 5 #, so 3-3=0 means in fact #... this means you have to code 0 somehow or not allowed 0 in your tree. This is how I would do it.

- p.andrey March 25, 2019 | Flag Reply
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0
of 0 vote

I think the idea with serialization is too complicated... I doubt this was what the interviewer wanted. But, the problem is easy... I didn't check all the corner cases, but this should be the solution (like Setu said... is just a xor between nodes, however the problem with id=0 remains):

public class Solution {
	
	private class Node {
		public Node left;
		public Node right;
		public int id ;
		
		public Node(int id) {
			this.left = null;
			this.right = null;
			this.id = id;
		}
	};

	private Node generateDiff(Node u, Node v) {
		if (u == null && v == null)
			return null;

		int id = 0;
		if (u != null)
			id = u.id;
		if (v != null)
			id ^= v.id;

		Node n = new Node(id);
		n.left = generateDiff(u == null ? null : u.left, v == null ? null : v.left);
		n.right = generateDiff(u == null ? null : u.right, v == null ? null : v.right);
		return n;
	}

	private Node generateBTfromDiff(Node d, Node u) {
		if (d == null && u == null)
			return null;

		int id = d.id;
		if (u != null)
			id ^= u.id;
		if (id == 0)
			return null;
		
		Node n = new Node(id);
		n.left = generateBTfromDiff(d.left, u == null ? null : u.left);
		n.right = generateBTfromDiff(d.right, u == null ? null : u.right);
		return n;
	}

	
	private Node[] buildBT() {
		Node [] roots = new Node[2];
		roots[0] = new Node(1);
		roots[0].right = new Node(3);
		roots[0].left = new Node(9);
		roots[0].right.left = new Node(8);

		roots[1] = new Node(2);
		roots[1].right = new Node(7);
		roots[1].left = new Node(4);
		roots[1].left.right = new Node(5);
		return roots;
	}
	
	private boolean checkBT(Node u, Node v) {
		if (u == null && v == null)
			return true;
		else if (u == null || v == null)
			return false;

		if (u.id != v.id)
			return false;
		
		return checkBT(u.left, v.left) && checkBT(u.right, v.right);
	}
	
	/* The MAIN ---------------------------------------------------------------------------------*/
	public static void main (String args[]) {
		BinaryTreeDifference t = new BinaryTreeDifference();
		Node roots[] = t.buildBT();
		Node d = t.generateDiff(roots[0], roots[1]);
		Node root_1 = t.generateBTfromDiff(d, roots[0]);
		Node root_0 = t.generateBTfromDiff(d, roots[1]);
		System.out.println("root 0 > Same ?  " + t.checkBT(root_0, roots[0]));
		System.out.println("root 1 > Same ?  " + t.checkBT(root_1, roots[1]));
	}
}

- p.andrey March 25, 2019 | Flag Reply
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0
of 0 vote

stackoverflow.com/questions/55244881/given-two-binary-trees-calculate-their-diff

- rubmeharderpls September 12, 2020 | Flag Reply


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