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Given an element x, task is to find the value of its immediate smaller element. Let res be the resultant node. Initialize the resultant Node as NULL. For every Node, check if data of root is greater than res, but less than x. if yes, update res. Recursively do the same for all nodes of the given Generic Tree. Return res, and res-&gt;key would be the immediate smaller element.Immediate Smaller element in an N-ary Tree: <a href="https://www.geeksforgeeks.org/immediate-smaller-element-n-ary-tree/">https://www.geeksforgeeks.org/immediate-smaller-element-n-ary-tree/</a>

Immediate Smaller element in an N-ary Tree

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. 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.Count number of nodes having more number of children than parents in n-ary tree : <a href="https://www.geeksforgeeks.org/given-n-ary-tree-count-number-nodes-number-children-parent/">https://www.geeksforgeeks.org/given-n-ary-tree-count-number-nodes-number-children-parent/</a>

Count number of nodes having more number of children than parents in n-ary tree

Given a generic tree and a integer x. Find and return the node with next larger element in the tree i.e. find a node just greater than x. Return NULL if no node is present with value greater than x.&nbsp; The idea is maintain a node pointer res, which will contain the final resultant node.&nbsp; Traverse the tree and check if root data is greater than x. If so, then compare the root data with res data.&nbsp; If root data is greater than n and less than res data update res. &nbsp;Next Larger element in n-ary tree : <a href="https://www.geeksforgeeks.org/next-larger-element-n-ary-tree/">https://www.geeksforgeeks.org/next-larger-element-n-ary-tree/</a>&nbsp;

Next Larger element in n-ary tree

The approach used is similar to Level Order traversal in a binary tree. Start by pushing the root node in the queue. And for each node, while popping it from queue, add the value of this node in the sum variable and push the children of the popped element in the queue. In case of a generic tree store child nodes in a vector. Thus, put all elements of the vector in the queue.Sum of all elements of N-ary Tree: <a href="https://www.geeksforgeeks.org/sum-elements-n-ary-tree/">https://www.geeksforgeeks.org/sum-elements-n-ary-tree/</a>

Sum of all elements of N-ary Tree

Given an N-Ary tree, find and return the node for which sum of data of all children and the node itself is maximum. In the sum, data of node itself and data of its immediate children is to be taken.The idea is we will maintain a integer variable maxsum which contains the maximum sum yet, and a resnode node pointer which points to the node with maximum sum.&nbsp; Traverse the tree and maintain the sum of root and data of all its immediate children in currsum&nbsp; integer variable and update the maxsum variable accordingly.Node having maximum sum of immediate children and itself in n-ary tree : <a href="https://www.geeksforgeeks.org/node-maximum-sum-immediate-children-n-ary-tree/">https://www.geeksforgeeks.org/node-maximum-sum-immediate-children-n-ary-tree/</a>

Node having maximum sum of immediate children and itself in n-ary tree

The idea is maintain a count variable initialize to 0. Traverse the tree and compare root data with x. If root data is greater than x, increment the count variable and recursively call for all its children.Number of nodes greater than a given value in n-ary tree: <a href="https://www.geeksforgeeks.org/number-nodes-greater-given-value-n-ary-tree/">https://www.geeksforgeeks.org/number-nodes-greater-given-value-n-ary-tree/</a>&nbsp;

Number of nodes greater than a given value in n-ary tree

Node of tree is represented as a key and a variable sized array of children pointers. The idea is similar to mirror of Binary Tree. For every node, we first recur for all of its children and then reverse array of children pointers. We can also do these steps in other way, i.e., reverse array of children pointers first and then recur for children. Below is C++ implementation of above idea.Mirror of n-ary Tree: <a href="https://www.geeksforgeeks.org/mirror-of-n-ary-tree/">https://www.geeksforgeeks.org/mirror-of-n-ary-tree/</a>

Mirror of n-ary Tree

Initialize the number of children as 0. For every node in the n-ary tree, check if its value is equal to x or not. If yes, then return the number of children. If the value of x is not equal to the current node then, push all the children of current node in the queue. Keep Repeating the above step until the queue becomes empty.Number of children of given node in n-ary Tree: <a href="https://www.geeksforgeeks.org/number-children-given-node-n-ary-tree/">https://www.geeksforgeeks.org/number-children-given-node-n-ary-tree/</a>

Number of children of given node in n-ary Tree

Given a generic tree, perform a Level order traversal and print all of its nodesThe approach to this problem is similar to Level Order traversal in a binary tree. We Start with pushing root node in a queue and for each node we pop it, print it and push all its child in the queue.In case of a generic tree we store child nodes in a vector. Thus we put all elements of the vector in the queue.&nbsp; Examples:&nbsp;Input : &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;10 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; / &nbsp; / &nbsp; &nbsp;\ &nbsp; \ &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;2 &nbsp;34 &nbsp; &nbsp;56 &nbsp; 100 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; / \ &nbsp; &nbsp; &nbsp; &nbsp;| &nbsp; / | \ &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;77 &nbsp;88 &nbsp; &nbsp; &nbsp;1 &nbsp; 7 &nbsp;8 &nbsp;9Output : 10 &nbsp; &nbsp; &nbsp; &nbsp; 2 34 56 100 &nbsp; &nbsp; &nbsp; &nbsp; 77 88 1 7 8 9General Tree Level Order Traversal : <a href="https://www.geeksforgeeks.org/generic-tree-level-order-traversal/">https://www.geeksforgeeks.org/generic-tree-level-order-traversal/</a>

General Tree Level Order Traversa

N-Ary tree can be traversed just like a normal tree. We just have to consider all childs of a given node and recursively call that function on every node.Depth of an N-Ary tree: <a href="https://www.geeksforgeeks.org/depth-n-ary-tree/">https://www.geeksforgeeks.org/depth-n-ary-tree/</a>&nbsp;

Depth of an N-Ary tree

Given an n-ary tree, count number of ways to traverse an n-ary (or a Directed Acyclic Graph) tree starting from the root vertex. Now we have to find the number of ways of traversing the whole tree starting from the root vertex. There can be many such ways. Some of them are listed below. 1) N-&gt;M-&gt;K-&gt;J-&gt;B-&gt;F-&gt;D-&gt;E-&gt;C-&gt;H-&gt;I-&gt;L-&gt;A (kind-of depth first traversal).&nbsp; 2) A-&gt;B-&gt;F-&gt;D-&gt;E-&gt;K-&gt;J-&gt;G-&gt;C-&gt;H-&gt;I-&gt;N-&gt;M-&gt;L (level order traversal)Number of ways to traverse an N-ary tree: <a href="https://www.geeksforgeeks.org/number-of-ways-to-traverse-an-n-ary-tree/">https://www.geeksforgeeks.org/number-of-ways-to-traverse-an-n-ary-tree/</a>

Number of ways to traverse an N-ary tree

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