Count of strictly increasing and decreasing paths in directed Binary Tree
Last Updated :
23 Jul, 2025
Given a directed binary tree of N nodes whose edges are directed from the parent to the child, the task is to count the number of strictly increasing and decreasing paths.
Note: A path starts at the root and ends at any leaf.
Examples:
Input: N = 6
tree = 6
/ \
4 7
\ / \
5 6 8
Output: 10 8
Explanation: For the above binary tree, the strictly increasing paths are :
All the paths consisting of a single node are = 6.
The paths containing two nodes are: 4->5, 7->8, 6->7 = 3.
The paths containing three nodes are: 6->7->8 = 1.
Hence, the count of strictly increasing paths is 6 + 3 + 1 = 10.
For the above binary tree, the strictly decreasing paths are:
All the paths consisting of a single node are = 6.
The paths containing two nodes are: 6->4, 7->6 = 2.
Hence, the count of strictly decreasing paths is 6 + 2 = 8.
Input: N = 5
tree 15
/
8
\
2
/ \
13 9
Output: 7 8
Explanation:
For the above binary tree, the strictly increasing paths are:
All the paths consisting of a single node are = 5.
The paths containing two nodes are: 2->9, 2->13 = 2.
Hence, the count of strictly increasing paths is 5 + 2 = 7.
For the above binary tree, the strictly decreasing paths are:
All the paths consisting of a single node are = 5.
The paths containing two nodes are:15->8, 8->2 = 2.
The paths containing three nodes are: 15->8->2 = 1.
Hence, the count of strictly decreasing paths is 5 + 2 + 1 = 8.
Approach: The problem can be solved based on the following observation:
If a path starting from any node u is strictly increasing and its value is greater than the value of its parent (say v), then all the strictly increasing paths starting from u are also strictly increasing paths when the starting node is v. The same is true in case of strictly decreasing paths.
From the above observation, the problem can be solved with the help of recursion. Follow the steps below to solve the problem:
- Use a recursive function to traverse the tree starting from the root and for each node return the count of strictly increasing and strictly decreasing paths from that node.
- In each recursion:
- Check if the path from the current node to a child (left or right) is increasing or decreasing.
- Call the recursive function for the child.
- Add the count of increasing or decreasing paths to the total count if the path from the current node to the child is increasing or decreasing respectively.
- Return the final count of increasing or decreasing path obtained for root.
Below is the implementation of the above approach :
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Following is the TreeNode
// class structure:
template <typename T>
class TreeNode {
public:
T data;
TreeNode<T>* left;
TreeNode<T>* right;
TreeNode(T data)
{
this->data = data;
left = NULL;
right = NULL;
}
};
// Helper function to get the answer
pair<int, int> countPathsHelper(TreeNode<int>* root,
int& increase,
int& decrease)
{
// If the root is NULL,
// then no strictly increasing or
// decreasing path.
if (root == NULL) {
return { 0, 0 };
}
// Call the function for both
// the left and right child.
pair<int, int> p1
= countPathsHelper(root->left,
increase, decrease);
pair<int, int> p2
= countPathsHelper(root->right,
increase, decrease);
// Initialize 'inc' and 'dec' to 1
int inc = 1, dec = 1;
// If the left child is not NULL.
if (root->left != NULL) {
// Check if the value is
// increasing from parent to child.
if (root->data < root->left->data) {
// Add the count of strictly
// increasing paths of
// child to parent.
inc += p1.first;
}
// Check if the value is decreasing
// from parent to child.
if (root->data > root->left->data) {
// Add the count of strictly
// decreasing paths of
// child to parent.
dec += p1.second;
}
}
if (root->right != NULL) {
// Check if the value is
// increasing from parent to child.
if (root->data < root->right->data) {
// Add the count of strictly
// increasing paths of
// child to parent.
inc += p2.first;
}
// Check if the value is
// decreasing from parent to child
if (root->data > root->right->data) {
// Add the count of strictly
// decreasing paths of
// child to parent.
dec += p2.second;
}
}
// Add the total count of
// strictly increasing paths to
// the global strictly increasing
// paths counter.
increase += inc;
// Add the total count of
// strictly decreasing paths to
// the global strictly
// decreasing paths counter.
decrease += dec;
return { inc, dec };
}
// Function to count the paths
pair<int, int> countPaths(TreeNode<int>* root)
{
// 'increase' stores the
// total strictly increasing paths.
int increase = 0;
// 'decrease' stores the
// total strictly decreasing paths.
int decrease = 0;
countPathsHelper(root, increase, decrease);
return { increase, decrease };
}
// Driver code
int main()
{
int N = 6;
TreeNode<int>* root
= new TreeNode<int>(N);
root->left = new TreeNode<int>(4);
root->right = new TreeNode<int>(7);
root->left->right = new TreeNode<int>(5);
root->right->left = new TreeNode<int>(6);
root->right->right = new TreeNode<int>(8);
// Function call
pair<int, int> ans = countPaths(root);
cout << ans.first << " " << ans.second;
return 0;
}
// Java code for the above approach:
import java.util.*;
public class Main {
static class Node {
int data;
Node left;
Node right;
Node(int data_value)
{
data = data_value;
this.left = null;
this.right = null;
}
}
// 'increase' stores the
// total strictly increasing paths.
static int increase;
// 'decrease' stores the
// total strictly decreasing paths.
static int decrease;
// Helper function to get the answer
static int[] countPathsHelper(Node root)
{
// If the root is NULL,
// then no strictly increasing or
// decreasing path.
if (root == null) {
int[] zero = {0,0};
return zero;
}
// Call the function for both
// the left and right child.
int[] p1
= countPathsHelper(root.left);
int[] p2
= countPathsHelper(root.right);
// Initialize 'inc' and 'dec' to 1
int inc = 1, dec = 1;
// If the left child is not NULL.
if (root.left != null) {
// Check if the value is
// increasing from parent to child.
if (root.data < root.left.data) {
// Add the count of strictly
// increasing paths of
// child to parent.
inc += p1[0];
}
// Check if the value is decreasing
// from parent to child.
if (root.data > root.left.data) {
// Add the count of strictly
// decreasing paths of
// child to parent.
dec += p1[1];
}
}
if (root.right != null) {
// Check if the value is
// increasing from parent to child.
if (root.data < root.right.data) {
// Add the count of strictly
// increasing paths of
// child to parent.
inc += p2[0];
}
// Check if the value is
// decreasing from parent to child
if (root.data > root.right.data) {
// Add the count of strictly
// decreasing paths of
// child to parent.
dec += p2[1];
}
}
// Add the total count of
// strictly increasing paths to
// the global strictly increasing
// paths counter.
increase += inc;
// Add the total count of
// strictly decreasing paths to
// the global strictly
// decreasing paths counter.
decrease += dec;
int[] ans = {inc,dec};
return ans;
}
// Function to count the paths
static int[] countPaths(Node root)
{
increase=0;
decrease=0;
countPathsHelper(root);
int[] ans = {increase,decrease};
return ans;
}
public static void main(String args[]) {
int N = 6;
Node root = new Node(N);
root.left = new Node(4);
root.right = new Node(7);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(8);
// Function call
int[] ans = countPaths(root);
System.out.println(ans[0] + " " + ans[1]);
}
}
// This code has been contributed by Sachin Sahara (sachin801)
# Python code for the above approach
## structure for a node
class Node:
def __init__(self, d):
self.data = d
self.left = None
self.right = None
## 'increase' stores the
## total strictly increasing paths.
increase = 0
## 'decrease' stores the
## total strictly decreasing paths.
decrease = 0
## Helper function to get the answer
def countPathsHelper(root):
global increase
global decrease
## If the root is NULL,
## then no strictly increasing or
## decreasing path.
if (root == None):
zero = [0, 0]
return zero
## Call the function for both
## the left and right child.
p1 = countPathsHelper(root.left)
p2 = countPathsHelper(root.right)
## Initialize 'inc' and 'dec' to 1
inc = 1
dec = 1
## If the left child is not NULL.
if (root.left != None):
## Check if the value is
## increasing from parent to child.
if (root.data < root.left.data):
## Add the count of strictly
## increasing paths of
## child to parent.
inc += p1[0];
## Check if the value is decreasing
## from parent to child.
if (root.data > root.left.data):
## Add the count of strictly
## decreasing paths of
## child to parent.
dec += p1[1]
if (root.right != None):
## Check if the value is
## increasing from parent to child.
if (root.data < root.right.data):
## Add the count of strictly
## increasing paths of
## child to parent.
inc += p2[0]
## Check if the value is
## decreasing from parent to child
if (root.data > root.right.data):
## Add the count of strictly
## decreasing paths of
## child to parent.
dec += p2[1]
## Add the total count of
## strictly increasing paths to
## the global strictly increasing
## paths counter.
increase += inc
## Add the total count of
## strictly decreasing paths to
## the global strictly
## decreasing paths counter.
decrease += dec
ans = [inc, dec]
return ans
## Function to count the paths
def countPaths(root):
global increase
global decrease
increase = 0
decrease = 0
countPathsHelper(root)
ans = [increase, decrease]
return ans
# Driver Code
if __name__=='__main__':
N = 6
root = Node(N)
root.left = Node(4)
root.right = Node(7)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(8)
## Function call
ans = countPaths(root)
print(ans[0], ans[1])
# This code is contributed by entertain2022.
// C# program to count the number of
// strictly increasing and decreasing paths
using System;
public class GFG{
public class Node {
public int data;
public Node left;
public Node right;
public Node(int data_value)
{
data = data_value;
this.left = null;
this.right = null;
}
}
// 'increase' stores the
// total strictly increasing paths.
static int increase;
// 'decrease' stores the
// total strictly decreasing paths.
static int decrease;
// Helper function to get the answer
static int[] countPathsHelper(Node root)
{
// If the root is NULL,
// then no strictly increasing or
// decreasing path.
if (root == null) {
int[] zero = {0,0};
return zero;
}
// Call the function for both
// the left and right child.
int[] p1 = countPathsHelper(root.left);
int[] p2 = countPathsHelper(root.right);
// Initialize 'inc' and 'dec' to 1
int inc = 1, dec = 1;
// If the left child is not NULL.
if (root.left != null) {
// Check if the value is
// increasing from parent to child.
if (root.data < root.left.data) {
// Add the count of strictly
// increasing paths of
// child to parent.
inc += p1[0];
}
// Check if the value is decreasing
// from parent to child.
if (root.data > root.left.data) {
// Add the count of strictly
// decreasing paths of
// child to parent.
dec += p1[1];
}
}
if (root.right != null) {
// Check if the value is
// increasing from parent to child.
if (root.data < root.right.data) {
// Add the count of strictly
// increasing paths of
// child to parent.
inc += p2[0];
}
// Check if the value is
// decreasing from parent to child
if (root.data > root.right.data) {
// Add the count of strictly
// decreasing paths of
// child to parent.
dec += p2[1];
}
}
// Add the total count of
// strictly increasing paths to
// the global strictly increasing
// paths counter.
increase += inc;
// Add the total count of
// strictly decreasing paths to
// the global strictly
// decreasing paths counter.
decrease += dec;
int[] ans = {inc,dec};
return ans;
}
// Function to count the paths
static int[] countPaths(Node root)
{
increase=0;
decrease=0;
countPathsHelper(root);
int[] ans = {increase,decrease};
return ans;
}
static public void Main (){
//Code
int N = 6;
Node root = new Node(N);
root.left = new Node(4);
root.right = new Node(7);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(8);
// Function call
int[] ans = countPaths(root);
Console.Write(ans[0] + " " + ans[1]);
}
}
// This code is contributed by shruti456rawal
// JavaScript code for the above approach
class TreeNode {
constructor(data) {
this.data = data;
this.left = null;
this.right = null;
}
}
function countPathsHelper(root, increase, decrease) {
if (root === null) {
return { inc: 0, dec: 0 };
}
const p1 = countPathsHelper(root.left, increase, decrease);
const p2 = countPathsHelper(root.right, increase, decrease);
let inc = 1;
let dec = 1;
if (root.left !== null) {
if (root.data < root.left.data) {
inc += p1.inc;
}
if (root.data > root.left.data) {
dec += p1.dec;
}
}
if (root.right !== null) {
if (root.data < root.right.data) {
inc += p2.inc;
}
if (root.data > root.right.data) {
dec += p2.dec;
}
}
increase.val += inc;
decrease.val += dec;
return { inc, dec };
}
function countPaths(root) {
const increase = { val: 0 };
const decrease = { val: 0 };
countPathsHelper(root, increase, decrease);
return { increase: increase.val, decrease: decrease.val };
}
const root = new TreeNode(6);
root.left = new TreeNode(4);
root.right = new TreeNode(7);
root.left.right = new TreeNode(5);
root.right.left = new TreeNode(6);
root.right.right = new TreeNode(8);
console.log(countPaths(root)['increase'] + " " + countPaths(root)['decrease']);
// This code is contributed by Potta Lokesh
Time Complexity: O(N)
Auxiliary Space: O(N)
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