# Preorder Traversal of N-ary Tree Without Recursion

Given an n-ary tree, print preorder traversal of it.

Example :

Preorder traversal of below tree is A B K N M J F D G E C H I L

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to use stack like iterative preorder traversal of binary tree.

1) Create an empty stack to store nodes.
2) Push the root node to the stack.
3) Run a loop while the stack is not empty
….a) Pop the top node from stack.
….b) Print the popped node.
….c) Store all the children of popped node onto the stack. We must store children from right to left so that leftmost node is popped first.
4) If stack is empty then we are done.

 `// C++ program to traverse an N-ary tree ` `// without recursion ` `#include ` `using` `namespace` `std; ` ` `  `// Structure of a node of an n-ary tree ` `struct` `Node { ` `    ``char` `key; ` `    ``vector child; ` `}; ` ` `  `// Utility function to create a new tree node ` `Node* newNode(``int` `key) ` `{ ` `    ``Node* temp = ``new` `Node; ` `    ``temp->key = key; ` `    ``return` `temp; ` `} ` ` `  `// Function to traverse tree without recursion ` `void` `traverse_tree(``struct` `Node* root) ` `{ ` `    ``// Stack to store the nodes ` `    ``stack nodes; ` ` `  `    ``// push the current node onto the stack ` `    ``nodes.push(root); ` ` `  `    ``// loop while the stack is not empty ` `    ``while` `(!nodes.empty()) { ` ` `  `        ``// store the current node and pop it from the stack ` `        ``Node* curr = nodes.top(); ` `        ``nodes.pop(); ` ` `  `        ``// current node has been travarsed ` `     ``if``(curr != NULL) ` `      ``{ ` `         ``cout << curr->key << ``" "``; ` ` `  `        ``// store all the childrent of current node from ` `        ``// right to left. ` `        ``vector::iterator it = curr->child.end(); ` ` `  `        ``while` `(it != curr->child.begin()) { ` `            ``it--; ` `            ``nodes.push(*it); ` `        ``} ` `      ``} ` `    ``} ` `} ` `// Driver program ` `int` `main() ` `{ ` `    ``/*   Let us create below tree  ` `   ``*            A  ` `   ``*        /  / \  \  ` `   ``*       B  F   D  E  ` `   ``*      / \     |  /|\  ` `   ``*     K  J     G C H I  ` `   ``*    / \         |   |  ` `   ``*   N   M        O   L  ` `   ``*/` ` `  `    ``Node* root = newNode(``'A'``); ` `    ``(root->child).push_back(newNode(``'B'``)); ` `    ``(root->child).push_back(newNode(``'F'``)); ` `    ``(root->child).push_back(newNode(``'D'``)); ` `    ``(root->child).push_back(newNode(``'E'``)); ` `    ``(root->child->child).push_back(newNode(``'K'``)); ` `    ``(root->child->child).push_back(newNode(``'J'``)); ` `    ``(root->child->child).push_back(newNode(``'G'``)); ` `    ``(root->child->child).push_back(newNode(``'C'``)); ` `    ``(root->child->child).push_back(newNode(``'H'``)); ` `    ``(root->child->child).push_back(newNode(``'I'``)); ` `    ``(root->child->child->child).push_back(newNode(``'N'``)); ` `    ``(root->child->child->child).push_back(newNode(``'M'``)); ` `    ``(root->child->child->child).push_back(newNode(``'O'``)); ` `    ``(root->child->child->child).push_back(newNode(``'L'``)); ` ` `  `    ``traverse_tree(root); ` ` `  `    ``return` `0; ` `} `

 `#Python program to find height of full binary tree  ` `# using preorder ` ` `  `class` `newNode():  ` `   `  `    ``def` `__init__(``self``, key):  ` `        ``self``.key ``=` `key ` `        ``# all children are stored in a list  ` `        ``self``.child ``=``[] ` `         `  `# Function to traverse tree without recursion  ` `def` `traverse_tree(root): ` `  `  `    ``# Stack to store the nodes  ` `    ``nodes``=``[] ` ` `  `    ``# push the current node onto the stack  ` `    ``nodes.append(root)  ` `   `  `    ``# loop while the stack is not empty  ` `    ``while` `(``len``(nodes)):   ` `   `  `        ``# store the current node and pop it from the stack  ` `        ``curr ``=` `nodes[``0``]  ` `        ``nodes.pop(``0``)  ` `   `  `        ``# current node has been travarsed  ` `        ``print``(curr.key,end``=``" "``) ` `        ``# store all the childrent of current node from  ` `        ``# right to left.  ` `        ``for` `it ``in` `range``(``len``(curr.child)``-``1``,``-``1``,``-``1``):   ` `            ``nodes.insert(``0``,curr.child[it]) ` `  `  `          `  `# Driver program to test above functions  ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``"""   Let us create below tree   ` `   ``*            A   ` `   ``*        /  / \  \   ` `   ``*       B  F   D  E   ` `   ``*      / \     |  /|\   ` `   ``*     K  J     G C H I   ` `   ``*    / \         |   |   ` `   ``*   N   M        O   L   ` `   ``"""` `    ``root ``=` `newNode(``'A'``)  ` `    ``(root.child).append(newNode(``'B'``))  ` `    ``(root.child).append(newNode(``'F'``))  ` `    ``(root.child).append(newNode(``'D'``))  ` `    ``(root.child).append(newNode(``'E'``))  ` `    ``(root.child[``0``].child).append(newNode(``'K'``))  ` `    ``(root.child[``0``].child).append(newNode(``'J'``))  ` `    ``(root.child[``2``].child).append(newNode(``'G'``))  ` `    ``(root.child[``3``].child).append(newNode(``'C'``))  ` `    ``(root.child[``3``].child).append(newNode(``'H'``))  ` `    ``(root.child[``3``].child).append(newNode(``'I'``))  ` `    ``(root.child[``0``].child[``0``].child).append(newNode(``'N'``))  ` `    ``(root.child[``0``].child[``0``].child).append(newNode(``'M'``))  ` `    ``(root.child[``3``].child[``0``].child).append(newNode(``'O'``))  ` `    ``(root.child[``3``].child[``2``].child).append(newNode(``'L'``))  ` `   `  `    ``traverse_tree(root) ` `  `  `# This code is contributed by SHUBHAMSINGH10 `

Output:
```A B K N M J F D G E C O H I L
```

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Improved By : SHUBHAMSINGH10, Enigma341

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