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)
- Traverse the tree in preorder or postorder (INORDER WON’T WORK)
- Maintain a stack where we maintain rolling sum with a rolling period as 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.
- 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:
Python3
# Python3 implementation of the approach# Class to store node informationclass Node: def __init__(self, val = None, left = None, right = None): self.val = val self.left = left self.right = right# Stack to perform stack operationsclass 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 = 0s = 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) |
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.
- Traverse the tree in preorder
- Maintain 3 variables called grandParent, parent, and child
- 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:
C++
// CPP implementation to print// the nodes having a single child#include <bits/stdc++.h>using namespace std;// Class of the Binary Tree nodestruct Node{ int data; Node *left, *right; Node(int x) { data = x; left = right = NULL; }};// global variableint 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 codeint 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 |
Java
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 |
Python
# Python3 program to implement# the above approachclass Node: def __init__(self, data): self.left = None self.right = None self.data = data # global variablecount = 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 codeif __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 |
C#
// 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 |
Output:
6
Time complexity: O(n)
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