Count all Grandparent-Parent-Child Triplets in a binary tree whose sum is greater than X

Given an integer X and a binary tree, the task is to count the number of triplet triplets of nodes such that their sum is greater than X and they have a grandparent -> parent -> child relationship.
 

Example:
 

Input: X = 100

               10
         /           \
       1              22
    /    \          /    \
  35      4       15       67
 /  \    /  \    /  \     /  \
57  38  9   10  110 312  131 414
                    /          \
                   8            39
Output: 6
The triplets are:
22 -> 15 -> 110
22 -> 15 -> 312
15 -> 312 -> 8
22 -> 67 -> 131
22 -> 67 -> 414
67 -> 414 -> 39

Approach: The problem can be solved using a rolling sum up approach with a rolling period as 3 (grandparent -> parent -> child) 

  1. Traverse the tree in preorder or postorder (INORDER WON’T WORK)
  2. Maintain a stack where we maintain rolling sum with a rolling period as 3
  3. Whenever we have more than 3 elements in the stack and if the topmost value is greater than X, we increment the result by 1.
  4. When we move up the recursion tree we do a POP operation on the stack so that all the rolling sums of lower levels get removed from the stack.

Below is the implementation of the above approach:
 



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# Python3 implementation of the approach
 
# Class to store node information
class Node:
    def __init__(self, val = None, left = None, right = None):
        self.val = val
        self.left = left
        self.right = right
 
# Stack to perform stack operations
class Stack:
    def __init__(self):
        self.stack = []
 
    @property
    def size(self):
        return len(self.stack)
 
    def top(self):
        return self.stack[self.size - 1] if self.size > 0 else 0
 
    def push(self, val):
        self.stack.append(val)
 
    def pop(self):
        if self.size >= 1:
            self.stack.pop(self.size - 1)
        else:
            self.stack = []
 
    # Period is 3 to satisfy grandparent-parent-child relationship
    def rolling_push(self, val, period = 3):
         
        # Find the index of element to remove
        to_remove_idx = self.size - period
 
        # If index is out of bounds then we remove nothing, i.e, 0
        to_remove = 0 if to_remove_idx < 0 else self.stack[to_remove_idx]
 
        # For rolling sum what we want is that at each index i,
        # we remove out-of-period elements, but since we are not
        # maintaining a separate list of actual elements,
        # we can get the actual element by taking diff between current
        # and previous element. So every time we remove an element,
        # we also add the element just before it, which is
        # equivalent to removing the actual value and not the rolling sum.
 
        # If index is out of bounds or 0 then we add nothing
        # i.e, 0, because there is no previous element
        to_add = 0 if to_remove_idx <= 0 else self.stack[to_remove_idx - 1]
 
        # If stack is empty then just push the value
        # Else add last element to current value to get rolling sum
        # then subtract out-of-period elements,
        # then finally add the element just before out-of-period element
        self.push(val if self.size <= 0 else val + self.stack[self.size - 1]
            - to_remove + to_add)
 
    def show(self):
        for item in self.stack:
            print(item)
 
 
# Global variables used by count_with_greater_sum()
count = 0
s = Stack()
 
def count_with_greater_sum(root_node, x):
    global s, count
 
    if not root_node:
        return 0
 
    s.rolling_push(root_node.val)
 
    if s.size >= 3 and s.top() > x:
        count += 1
 
    count_with_greater_sum(root_node.left, x)
    count_with_greater_sum(root_node.right, x)
 
    # Moving up the tree so pop the last element
    s.pop()
 
 
if __name__ == '__main__':
    root = Node(10)
 
    root.left = Node(1)
    root.right = Node(22)
 
    root.left.left = Node(35)
    root.left.right = Node(4)
 
    root.right.left = Node(15)
    root.right.right = Node(67)
 
    root.left.left.left = Node(57)
    root.left.left.right = Node(38)
 
    root.left.right.left = Node(9)
    root.left.right.right = Node(10)
 
    root.right.left.left = Node(110)
    root.right.left.right = Node(312)
 
    root.right.right.left = Node(131)
    root.right.right.right = Node(414)
 
    root.right.left.right.left = Node(8)
 
    root.right.right.right.right = Node(39)
 
    count_with_greater_sum(root, 100)
 
    print(count)
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Output:
6






 

Efficient Approach: The problem can be solved by maintaining 3 variables called grandparent, parent, and child. It can be done in constant space without using other data structures. 

  1. Traverse the tree in preorder
  2. Maintain 3 variables called grandParent, parent, and child
  3. Whenever we have sum more than the target we can increase the count or print the triplet.

Below is the implementation of the above approach: 

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// CPP implementation to print
// the nodes having a single child
#include <bits/stdc++.h>
using namespace std;
 
// Class of the Binary Tree node
struct Node
{
  int data;
  Node *left, *right;
 
  Node(int x)
  {
    data = x;
    left = right = NULL;
  }
};
 
// global variable
int cont = 0;
 
void preorder(Node* grandParent,
              Node* parent,
              Node* child,
              int sum)
{
  if(grandParent != NULL &&
     parent != NULL &&
     child != NULL &&
     (grandParent -> data +
      parent -> data +
      child->data) > sum)
  {
    cont++;
    //uncomment below lines if you
    // want to print triplets
    /*System->out->print(grandParent
      ->data+"-->"+parent->data+"-->
      "+child->data);
      System->out->println();*/
  }
  if(child == NULL)
    return;
   
  preorder(parent, child,
           child -> left, sum);
  preorder(parent, child,
           child -> right, sum);
}
 
//Driver code
int main()
{
  Node *r10 = new Node(10);
  Node *r1 = new Node(1);
  Node *r22 = new Node(22);
  Node *r35 = new Node(35);
  Node *r4 = new Node(4);
  Node *r15 = new Node(15);
  Node *r67 = new Node(67);
  Node *r57 = new Node(57);
  Node *r38 = new Node(38);
  Node *r9 = new Node(9);
  Node *r10_2 = new Node(10);
  Node *r110 = new Node(110);
  Node *r312 = new Node(312);
  Node *r131 = new Node(131);
  Node *r414 = new Node(414);
  Node *r8 = new Node(8);
  Node *r39 = new Node(39);
 
  r10 -> left = r1;
  r10 -> right = r22;
  r1 -> left = r35;
  r1 -> right = r4;
  r22 -> left = r15;
  r22 -> right = r67;
  r35 -> left = r57;
  r35 -> right = r38;
  r4 -> left = r9;
  r4 -> right = r10_2;
  r15 -> left = r110;
  r15 -> right = r312;
  r67 -> left = r131;
  r67 -> right = r414;
  r312 -> left = r8;
  r414 -> right = r39;
 
  preorder(NULL, NULL,
           r10, 100);
  cout << cont;
}
 
// This code is contributed by Mohit Kumar 29
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class Node{
    int data;
    Node left;
    Node right;
    public Node(int data)
    {
        this.data=data;
    }
}
 class TreeTriplet {
    static int count=0; // global variable
    public void preorder(Node grandParent,Node parent,Node child,int sum)
    {
        if(grandParent!=null && parent!=null && child!=null && (grandParent.data+parent.data+child.data) > sum)
        {
            count++;
            //uncomment below lines if you want to print triplets
            /*System.out.print(grandParent.data+"-->"+parent.data+"-->"+child.data);
            System.out.println();*/           
        }
        if(child==null)
          return;       
        preorder(parent,child,child.left,sum);
        preorder(parent,child,child.right,sum);       
    }
    public static void main(String args[])
    {       
        Node r10 = new Node(10);
        Node r1 = new Node(1);
        Node r22 = new Node(22);
        Node r35 = new Node(35);
        Node r4 = new Node(4);
        Node r15 = new Node(15);
        Node r67 = new Node(67);
        Node r57 = new Node(57);
        Node r38 = new Node(38);
        Node r9 = new Node(9);
        Node r10_2 = new Node(10);
        Node r110 = new Node(110);
        Node r312 = new Node(312);
        Node r131 = new Node(131);
        Node r414 = new Node(414);
        Node r8 = new Node(8);
        Node r39 = new Node(39);
         
        r10.left=r1;
        r10.right=r22;
        r1.left=r35;
        r1.right=r4;
        r22.left=r15;
        r22.right=r67;
        r35.left=r57;
        r35.right=r38;
        r4.left=r9;
        r4.right=r10_2;
        r15.left=r110;
        r15.right=r312;
        r67.left=r131;
        r67.right=r414;
        r312.left=r8;
        r414.right=r39;   
 
        TreeTriplet p = new TreeTriplet();
        p.preorder(null, null, r10,100);
        System.out.println(count);
}
     
     
}
 // This code is contributed by Akshay Siddhpura
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# Python3 program to implement
# the above approach
class Node:
     
    def __init__(self,
                 data):
         
        self.left = None
        self.right = None
        self.data = data       
 
# global variable
count = 0
     
def preorder(grandParent, parent,
             child, sum):
     
    global count
         
    if(grandParent != None and
       parent != None and
       child != None and
       (grandParent.data +
        parent.data +
        child.data) > sum):    
        count += 1
             
        # uncomment below lines if
        # you want to print triplets 
        # System.out.print(grandParent.
        # data+"-->"+parent.data+"-->
        # "+child.data); System.out.println();          
    if(child == None):
        return;
           
    preorder(parent, child,
             child.left, sum);
    preorder(parent, child,
             child.right, sum); 
     
# Driver code
if __name__ == "__main__":
     
    r10 = Node(10);
    r1 = Node(1);
    r22 = Node(22);
    r35 = Node(35);
    r4 = Node(4);
    r15 = Node(15);
    r67 = Node(67);
    r57 = Node(57);
    r38 = Node(38);
    r9 = Node(9);
    r10_2 = Node(10);
    r110 = Node(110);
    r312 = Node(312);
    r131 = Node(131);
    r414 = Node(414);
    r8 = Node(8);
    r39 = Node(39);
           
    r10.left = r1;
    r10.right = r22;
    r1.left = r35;
    r1.right = r4;
    r22.left = r15;
    r22.right = r67;
    r35.left = r57;
    r35.right = r38;
    r4.left = r9;
    r4.right = r10_2;
    r15.left = r110;
    r15.right = r312;
    r67.left = r131;
    r67.right = r414;
    r312.left = r8;
    r414.right = r39;    
   
    preorder(None, None,
             r10, 100)
    print(count);
         
# This code is contributed by Rutvik_56
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// C# program to find an index which has
// same number of even elements on left and
// right, Or same number of odd elements on
// left and right.
using System;
     
public class Node
{
    public int data;
    public Node left;
    public Node right;
    public Node(int data)
    {
        this.data = data;
    }
}
 
class GFG
{
    static int count = 0; // global variable
    public void preorder(Node grandParent,
                         Node parent,
                         Node child, int sum)
    {
        if(grandParent != null && parent != null &&
                 child != null && (grandParent.data +
                     parent.data + child.data) > sum)
        {
            count++;
             
            // uncomment below lines if you want to print triplets
            /*System.out.print(grandParent.data+"-->"+parent.data+"-->"+child.data);
            System.out.println();*/       
        }
        if(child == null)
            return;    
        preorder(parent,child,child.left,sum);
        preorder(parent,child,child.right,sum);    
    }
     
    // Driver Code
    public static void Main(String []args)
    {    
        Node r10 = new Node(10);
        Node r1 = new Node(1);
        Node r22 = new Node(22);
        Node r35 = new Node(35);
        Node r4 = new Node(4);
        Node r15 = new Node(15);
        Node r67 = new Node(67);
        Node r57 = new Node(57);
        Node r38 = new Node(38);
        Node r9 = new Node(9);
        Node r10_2 = new Node(10);
        Node r110 = new Node(110);
        Node r312 = new Node(312);
        Node r131 = new Node(131);
        Node r414 = new Node(414);
        Node r8 = new Node(8);
        Node r39 = new Node(39);
         
        r10.left = r1;
        r10.right = r22;
        r1.left = r35;
        r1.right = r4;
        r22.left = r15;
        r22.right = r67;
        r35.left = r57;
        r35.right = r38;
        r4.left = r9;
        r4.right = r10_2;
        r15.left = r110;
        r15.right = r312;
        r67.left = r131;
        r67.right = r414;
        r312.left = r8;
        r414.right = r39;
 
        GFG p = new GFG();
        p.preorder(null, null, r10,100);
        Console.WriteLine(count);
}
}
 
// This code is contributed by 29AjayKumar
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Output:
6






 

Time complexity: O(n) 

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