Jobs

Interviews

Discussions

N-ary Tree Postorder Traversal

#930 Most AskedEasy

N-ary Tree Postorder Traversal #930 Most Asked

Easy

Bloomberg LP

8 views

Topics:

TreesRecursion

Given the root of an n-ary tree, return the postorder traversal of its nodes' values.

Nary-Tree input serialization is represented in their level order traversal. Each group of children is separated by the null value (See examples)

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [5,6,3,2,4,1]

Example 2:

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]

Constraints:

The number of nodes in the tree is in the range [0, 10⁴].
0 <= Node.val <= 10⁴
The height of the n-ary tree is less than or equal to 1000.

Follow up: Recursive solution is trivial, could you do it iteratively?

Try coding it in LeetCode

N-ary Tree Postorder Traversal

#930 Most AskedEasy

N-ary Tree Postorder Traversal #930 Most Asked

Easy

Bloomberg LP

8 views

Topics:

TreesRecursion

Given the root of an n-ary tree, return the postorder traversal of its nodes' values.

Nary-Tree input serialization is represented in their level order traversal. Each group of children is separated by the null value (See examples)

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [5,6,3,2,4,1]

Example 2:

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]

Constraints:

The number of nodes in the tree is in the range [0, 10⁴].
0 <= Node.val <= 10⁴
The height of the n-ary tree is less than or equal to 1000.

Follow up: Recursive solution is trivial, could you do it iteratively?

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the 'val' attribute of each node in the N-ary tree?
Can the N-ary tree be empty (null or contain no nodes)? What should I return in that case?
What is the maximum number of children a node can have (the 'children' list's maximum size)?
Is the order of children within a node's 'children' list significant for the postorder traversal, or can I assume it's arbitrary?
Should I return the list of node values as integers or strings?

Brute Force Solution

Approach

The brute force approach to traversing a tree in postorder means we want to visit all the children of a node before visiting the node itself. To do this the brute force way, we will explore all possibilities of visiting the children first, one by one.

Here's how the algorithm would work step-by-step:

Start with the root of the tree.
For each child of the root, completely traverse that child's entire subtree in postorder.
To traverse a subtree in postorder, repeat the above step for each child of the current node in that subtree.
Once you've traversed all the children of a particular node, then and only then visit that node itself.
Continue this process until every node in the entire tree has been visited.

Code Implementation

def n_ary_tree_postorder_traversal(root):
    results = []

    def traverse_node(node):
        if not node:
            return

        # Traverse each child's subtree before visiting the node
        if node.children:

            for child_node in node.children:

                traverse_node(child_node)

        # Append the node's value after visiting all children
        results.append(node.val)

    traverse_node(root)

    return results

Big(O) Analysis

Time Complexity

O(n) – The provided traversal visits each node in the n-ary tree exactly once. The algorithm processes each node by recursively visiting all of its children before visiting the node itself. Since each node is visited and processed a constant amount of time (specifically, only once), the total time complexity is directly proportional to the number of nodes in the tree. Therefore, the time complexity is O(n), where n is the number of nodes.

Space Complexity

O(H) – The space complexity is determined by the maximum depth of the recursion stack. In the worst case, where the tree resembles a linked list (highly skewed), the depth can be proportional to the total number of nodes, N. However, more generally, the maximum depth will be equivalent to the height (H) of the n-ary tree, since the algorithm will have a stack frame for each level of the tree during the recursive calls. Therefore, the auxiliary space complexity is O(H), where H is the height of the N-ary tree. In the worst case where H is equal to N, the space complexity simplifies to O(N).

Optimal Solution

Approach

To traverse an N-ary tree in postorder, we need to visit all the children of a node before visiting the node itself. A good way to do this is by using a temporary storage space (like a to-do list) to keep track of nodes we still need to visit.

Here's how the algorithm would work step-by-step:

Start with the root node of the tree.
Put the root node in your temporary storage space.
While your temporary storage space is not empty, take the last node you added from it.
Check if the node has any children that you have not visited yet.
If it has unvisited children, add all those children to your temporary storage space, making sure to add the leftmost child last.
If the node has no unvisited children, it means you have visited all its children. Add the node's value to your final result list.
Repeat steps 3-6 until your temporary storage space is empty.
The final result list will contain the values of the nodes in postorder.

Code Implementation

class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children

def postorder_traversal(root):
    if not root:
        return []

    result_list = []
    nodes_to_visit = [root]

    while nodes_to_visit:
        current_node = nodes_to_visit.pop()

        # Check if the current node has unvisited children
        if current_node.children:
            children = current_node.children
            nodes_to_visit.append(current_node)
            # Add children in reverse order for correct processing.
            for child in reversed(children):
                nodes_to_visit.append(child)
        else:
            # Add node to the result because its children are visited
            result_list.append(current_node.val)

    return result_list

Big(O) Analysis

Time Complexity

O(n) – The algorithm visits each node in the N-ary tree exactly once. In the worst-case scenario, the while loop iterates through all n nodes of the tree. The operations inside the loop (checking for children and adding to/removing from the stack and result list) take constant time. Therefore, the overall time complexity is O(n), where n is the number of nodes in the N-ary tree.

Space Complexity

O(N) – The plain English explanation uses a temporary storage space, which effectively acts as a stack, to store nodes. In the worst-case scenario, where the N-ary tree resembles a single branch, the temporary storage space could hold all N nodes of the tree. Therefore, the auxiliary space used by the algorithm is proportional to the number of nodes, N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty root node	Return an empty list if the root is null or empty, indicating an empty tree.
Root node with no children	Return a list containing only the root node's value, as it's the only node to traverse.
Very deep tree (high recursion depth)	Consider using an iterative approach (stack-based) to avoid potential stack overflow errors.
Tree with a very large number of nodes (memory constraints)	Ensure the chosen data structures (like the list to store the result) can handle the large number of node values without excessive memory usage.
Tree with a very wide branching factor (many children per node)	The solution should efficiently iterate through all the children of each node.
Integer overflow in node values (if applicable)	If node values are integers, consider the possibility of overflow during calculations (if any are needed beyond simple traversal) and use appropriate data types or checks.
Tree with duplicate node values	The traversal order is not affected by duplicate values, so the algorithm should produce a correct postorder traversal regardless.
N-ary tree with disconnected subtrees (not a single connected component from the root)	The algorithm assumes the input is a single, connected N-ary tree stemming from the root; disconnected subtrees will not be traversed.

Null or empty root node

How to Handle:

Return an empty list if the root is null or empty, indicating an empty tree.

Root node with no children

How to Handle:

Return a list containing only the root node's value, as it's the only node to traverse.

Very deep tree (high recursion depth)

How to Handle:

Consider using an iterative approach (stack-based) to avoid potential stack overflow errors.

Tree with a very large number of nodes (memory constraints)

How to Handle:

Ensure the chosen data structures (like the list to store the result) can handle the large number of node values without excessive memory usage.

Tree with a very wide branching factor (many children per node)

How to Handle:

The solution should efficiently iterate through all the children of each node.

Integer overflow in node values (if applicable)

How to Handle:

If node values are integers, consider the possibility of overflow during calculations (if any are needed beyond simple traversal) and use appropriate data types or checks.

Tree with duplicate node values

How to Handle:

The traversal order is not affected by duplicate values, so the algorithm should produce a correct postorder traversal regardless.

N-ary tree with disconnected subtrees (not a single connected component from the root)

How to Handle:

The algorithm assumes the input is a single, connected N-ary tree stemming from the root; disconnected subtrees will not be traversed.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the 'val' attribute of each node in the N-ary tree?
Can the N-ary tree be empty (null or contain no nodes)? What should I return in that case?
What is the maximum number of children a node can have (the 'children' list's maximum size)?
Is the order of children within a node's 'children' list significant for the postorder traversal, or can I assume it's arbitrary?
Should I return the list of node values as integers or strings?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the root of the tree.
For each child of the root, completely traverse that child's entire subtree in postorder.
To traverse a subtree in postorder, repeat the above step for each child of the current node in that subtree.
Once you've traversed all the children of a particular node, then and only then visit that node itself.
Continue this process until every node in the entire tree has been visited.

Code Implementation

def n_ary_tree_postorder_traversal(root):
    results = []

    def traverse_node(node):
        if not node:
            return

        # Traverse each child's subtree before visiting the node
        if node.children:

            for child_node in node.children:

                traverse_node(child_node)

        # Append the node's value after visiting all children
        results.append(node.val)

    traverse_node(root)

    return results

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the root node of the tree.
Put the root node in your temporary storage space.
While your temporary storage space is not empty, take the last node you added from it.
Check if the node has any children that you have not visited yet.
If it has unvisited children, add all those children to your temporary storage space, making sure to add the leftmost child last.
If the node has no unvisited children, it means you have visited all its children. Add the node's value to your final result list.
Repeat steps 3-6 until your temporary storage space is empty.
The final result list will contain the values of the nodes in postorder.

Code Implementation

class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children

def postorder_traversal(root):
    if not root:
        return []

    result_list = []
    nodes_to_visit = [root]

    while nodes_to_visit:
        current_node = nodes_to_visit.pop()

        # Check if the current node has unvisited children
        if current_node.children:
            children = current_node.children
            nodes_to_visit.append(current_node)
            # Add children in reverse order for correct processing.
            for child in reversed(children):
                nodes_to_visit.append(child)
        else:
            # Add node to the result because its children are visited
            result_list.append(current_node.val)

    return result_list

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty root node	Return an empty list if the root is null or empty, indicating an empty tree.
Root node with no children	Return a list containing only the root node's value, as it's the only node to traverse.
Very deep tree (high recursion depth)	Consider using an iterative approach (stack-based) to avoid potential stack overflow errors.
Tree with a very large number of nodes (memory constraints)	Ensure the chosen data structures (like the list to store the result) can handle the large number of node values without excessive memory usage.
Tree with a very wide branching factor (many children per node)	The solution should efficiently iterate through all the children of each node.
Integer overflow in node values (if applicable)	If node values are integers, consider the possibility of overflow during calculations (if any are needed beyond simple traversal) and use appropriate data types or checks.
Tree with duplicate node values	The traversal order is not affected by duplicate values, so the algorithm should produce a correct postorder traversal regardless.
N-ary tree with disconnected subtrees (not a single connected component from the root)	The algorithm assumes the input is a single, connected N-ary tree stemming from the root; disconnected subtrees will not be traversed.

Null or empty root node

How to Handle:

Return an empty list if the root is null or empty, indicating an empty tree.

Root node with no children

How to Handle:

Return a list containing only the root node's value, as it's the only node to traverse.

Very deep tree (high recursion depth)

How to Handle:

Consider using an iterative approach (stack-based) to avoid potential stack overflow errors.

Tree with a very large number of nodes (memory constraints)

How to Handle:

Ensure the chosen data structures (like the list to store the result) can handle the large number of node values without excessive memory usage.

Tree with a very wide branching factor (many children per node)

How to Handle:

The solution should efficiently iterate through all the children of each node.

Integer overflow in node values (if applicable)

How to Handle:

If node values are integers, consider the possibility of overflow during calculations (if any are needed beyond simple traversal) and use appropriate data types or checks.

Tree with duplicate node values

How to Handle:

The traversal order is not affected by duplicate values, so the algorithm should produce a correct postorder traversal regardless.

N-ary tree with disconnected subtrees (not a single connected component from the root)

How to Handle:

The algorithm assumes the input is a single, connected N-ary tree stemming from the root; disconnected subtrees will not be traversed.

0/1037 Bloomberg LP questions completed

0/1037 completed