Given a N-ary tree represented as adjacency list, we need to write a program to count all such nodes in this tree which has more number of children than its parent.

For Example,

In the above tree, the count will be 1 as there is only one such node which is ‘2’ which has more number of children than its parent. 2 has three children (4, 5 and 6) whereas its parent, 1 has only two children (2 and 3).

We can solve this problem using both BFS and DFS algorithms. We will explain here in details about how to solve this problem using BFS algorithm.

As the tree is represented using adjacency list representation. So, for any node say ‘u’ the number of children of this node can be given as adj[u].size().

Now the idea is to apply BFS on the given tree and while traversing the children of a node ‘u’ say ‘v’ we will simply check is adj[v].size() > adj[u].size().

Below is the implementation of above idea:

## CPP

`// C++ program to count number of nodes ` `// which has more children than its parent ` ` ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to count number of nodes ` `// which has more children than its parent ` `int` `countNodes(vector<` `int` `> adj[], ` `int` `root) ` `{ ` ` ` `int` `count = 0; ` ` ` ` ` `// queue for applying BFS ` ` ` `queue<` `int` `> q; ` ` ` ` ` `// BFS algorithm ` ` ` `q.push(root); ` ` ` ` ` `while` `(!q.empty()) ` ` ` `{ ` ` ` `int` `node = q.front(); ` ` ` `q.pop(); ` ` ` ` ` `// traverse children of node ` ` ` `for` `( ` `int` `i=0;i<adj[node].size();i++) ` ` ` `{ ` ` ` `// children of node ` ` ` `int` `children = adj[node][i]; ` ` ` ` ` `// if number of childs of children ` ` ` `// is greater than number of childs ` ` ` `// of node, then increment count ` ` ` `if` `(adj[children].size() > adj[node].size()) ` ` ` `count++; ` ` ` `q.push(children); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `count; ` `} ` ` ` `// Driver program to test above functions ` `int` `main() ` `{ ` ` ` `// adjacency list for n-ary tree ` ` ` `vector<` `int` `> adj[10]; ` ` ` ` ` `// construct n ary tree as shown ` ` ` `// in above diagram ` ` ` `adj[1].push_back(2); ` ` ` `adj[1].push_back(3); ` ` ` `adj[2].push_back(4); ` ` ` `adj[2].push_back(5); ` ` ` `adj[2].push_back(6); ` ` ` `adj[3].push_back(9); ` ` ` `adj[5].push_back(7); ` ` ` `adj[5].push_back(8); ` ` ` ` ` `int` `root = 1; ` ` ` ` ` `cout << countNodes(adj, root); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to count number of nodes ` `// which has more children than its parent ` `import` `java.util.*; ` ` ` `class` `GFG ` `{ ` ` ` `// function to count number of nodes ` `// which has more children than its parent ` `static` `int` `countNodes(Vector<Integer> adj[], ` `int` `root) ` `{ ` ` ` `int` `count = ` `0` `; ` ` ` ` ` `// queue for applying BFS ` ` ` `Queue<Integer> q = ` `new` `LinkedList<>(); ` ` ` ` ` `// BFS algorithm ` ` ` `q.add(root); ` ` ` ` ` `while` `(!q.isEmpty()) ` ` ` `{ ` ` ` `int` `node = q.peek(); ` ` ` `q.remove(); ` ` ` ` ` `// traverse children of node ` ` ` `for` `( ` `int` `i=` `0` `;i<adj[node].size();i++) ` ` ` `{ ` ` ` `// children of node ` ` ` `int` `children = adj[node].get(i); ` ` ` ` ` `// if number of childs of children ` ` ` `// is greater than number of childs ` ` ` `// of node, then increment count ` ` ` `if` `(adj[children].size() > adj[node].size()) ` ` ` `count++; ` ` ` `q.add(children); ` ` ` `} ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `// adjacency list for n-array tree ` ` ` `Vector<Integer> []adj = ` `new` `Vector[` `10` `]; ` ` ` `for` `(` `int` `i= ` `0` `; i < ` `10` `; i++) { ` ` ` `adj[i] = ` `new` `Vector<>(); ` ` ` `} ` ` ` ` ` `// conn array tree as shown ` ` ` `// in above diagram ` ` ` `adj[` `1` `].add(` `2` `); ` ` ` `adj[` `1` `].add(` `3` `); ` ` ` `adj[` `2` `].add(` `4` `); ` ` ` `adj[` `2` `].add(` `5` `); ` ` ` `adj[` `2` `].add(` `6` `); ` ` ` `adj[` `3` `].add(` `9` `); ` ` ` `adj[` `5` `].add(` `7` `); ` ` ` `adj[` `5` `].add(` `8` `); ` ` ` ` ` `int` `root = ` `1` `; ` ` ` ` ` `System.out.print(countNodes(adj, root)); ` `} ` `} ` ` ` `// This code is contributed by Rajput-Ji ` |

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## Python3

`# Python3 program to count number of nodes ` `# which has more children than its parent ` `from` `collections ` `import` `deque ` ` ` `adj ` `=` `[[] ` `for` `i ` `in` `range` `(` `100` `)] ` ` ` `# function to count number of nodes ` `# which has more children than its parent ` `def` `countNodes(root): ` ` ` ` ` `count ` `=` `0` ` ` ` ` `# queue for applying BFS ` ` ` `q ` `=` `deque() ` ` ` ` ` `# BFS algorithm ` ` ` `q.append(root) ` ` ` ` ` `while` `len` `(q) > ` `0` `: ` ` ` ` ` `node ` `=` `q.popleft() ` ` ` ` ` `# traverse children of node ` ` ` `for` `i ` `in` `adj[node]: ` ` ` `# children of node ` ` ` `children ` `=` `i ` ` ` ` ` `# if number of childs of children ` ` ` `# is greater than number of childs ` ` ` `# of node, then increment count ` ` ` `if` `(` `len` `(adj[children]) > ` `len` `(adj[node])): ` ` ` `count ` `+` `=` `1` ` ` `q.append(children) ` ` ` ` ` `return` `count ` ` ` ` ` `# Driver program to test above functions ` ` ` `# construct n ary tree as shown ` `# in above diagram ` `adj[` `1` `].append(` `2` `) ` `adj[` `1` `].append(` `3` `) ` `adj[` `2` `].append(` `4` `) ` `adj[` `2` `].append(` `5` `) ` `adj[` `2` `].append(` `6` `) ` `adj[` `3` `].append(` `9` `) ` `adj[` `5` `].append(` `7` `) ` `adj[` `5` `].append(` `8` `) ` ` ` `root ` `=` `1` ` ` `print` `(countNodes(root)) ` ` ` `# This code is contributed by mohit kumar 29 ` |

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## C#

`// C# program to count number of nodes ` `// which has more children than its parent ` `using` `System; ` `using` `System.Collections.Generic; ` ` ` `class` `GFG ` `{ ` ` ` `// function to count number of nodes ` `// which has more children than its parent ` `static` `int` `countNodes(List<` `int` `> []adj, ` `int` `root) ` `{ ` ` ` `int` `count = 0; ` ` ` ` ` `// queue for applying BFS ` ` ` `List<` `int` `> q = ` `new` `List<` `int` `>(); ` ` ` ` ` `// BFS algorithm ` ` ` `q.Add(root); ` ` ` ` ` `while` `(q.Count != 0) ` ` ` `{ ` ` ` `int` `node = q[0]; ` ` ` `q.RemoveAt(0); ` ` ` ` ` `// traverse children of node ` ` ` `for` `( ` `int` `i = 0; i < adj[node].Count; i++) ` ` ` `{ ` ` ` `// children of node ` ` ` `int` `children = adj[node][i]; ` ` ` ` ` `// if number of childs of children ` ` ` `// is greater than number of childs ` ` ` `// of node, then increment count ` ` ` `if` `(adj[children].Count > adj[node].Count) ` ` ` `count++; ` ` ` `q.Add(children); ` ` ` `} ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `// adjacency list for n-array tree ` ` ` `List<` `int` `> []adj = ` `new` `List<` `int` `>[10]; ` ` ` `for` `(` `int` `i= 0; i < 10 ; i++) { ` ` ` `adj[i] = ` `new` `List<` `int` `>(); ` ` ` `} ` ` ` ` ` `// conn array tree as shown ` ` ` `// in above diagram ` ` ` `adj[1].Add(2); ` ` ` `adj[1].Add(3); ` ` ` `adj[2].Add(4); ` ` ` `adj[2].Add(5); ` ` ` `adj[2].Add(6); ` ` ` `adj[3].Add(9); ` ` ` `adj[5].Add(7); ` ` ` `adj[5].Add(8); ` ` ` ` ` `int` `root = 1; ` ` ` ` ` `Console.Write(countNodes(adj, root)); ` `} ` `} ` ` ` `// This code is contributed by PrinciRaj1992 ` |

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**Output:**

1

**Time Complexity**: O( n ) , where n is the number of nodes in the tree.

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