#590

Easy

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

N-ary Tree Postorder Traversal

Postorder traversal of an N-ary tree can be efficiently solved using DFS and stack-based methods, tracking state across nodes.

stack tree depth first search

Problem Statement

Given the root of an N-ary tree, return the postorder traversal of its nodes' values. In postorder traversal, the nodes are visited in the order: left, right, root, with the traversal occurring from bottom to top. N-ary trees are trees where each node can have any number of children, as opposed to the binary tree where each node has at most two children.

The input is provided in level-order, where each node is followed by its children, and null is used to separate different children for each node. The challenge here is to perform DFS while keeping track of the correct traversal order using stack-based techniques.

Examples

Example 1

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

Output: [5,6,3,2,4,1]

Example details omitted.

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]

Example details omitted.

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.

Solution Approach

DFS with Stack

Start with a DFS approach, using a stack to manage nodes. Push the root node onto the stack, and traverse each node's children by pushing them onto the stack in reverse order. Once a node’s children are processed, visit the node itself.

Iterative Approach

This problem can be solved iteratively using a stack for efficient management of nodes. Keep track of the state for each node and ensure that children are traversed before their parents by processing them in the reverse order.

State Management with Stack

Proper state management is critical. Each time a child is visited, track it and mark it as processed, ensuring the parent is visited only after all children are processed. This ensures the postorder pattern is maintained.

Complexity Analysis

Metric	Value
Time	O(m)
Space	O(m)

Both time and space complexity are O(m), where m is the total number of nodes in the tree. This complexity arises because each node is processed once, and the stack stores at most m nodes during traversal.

What Interviewers Usually Probe

Ability to manage state during tree traversal
Skill in implementing DFS using stack
Efficiency in handling large trees with many nodes

Common Pitfalls or Variants

Common pitfalls

Not managing state correctly, leading to incorrect traversal order
Pushing children onto the stack in the wrong order
Missing edge case handling for an empty tree

Follow-up variants

Iterative solution using DFS with explicit stack management
Recursion-based solution for postorder traversal
Optimized DFS for large N-ary trees with more than 1000 nodes

FAQ

How do you handle large N-ary trees in postorder traversal?

By using DFS and a stack, managing state efficiently, and ensuring the tree is processed iteratively to handle large inputs effectively.

What is the time complexity of the N-ary Tree Postorder Traversal?

The time complexity is O(m), where m is the number of nodes in the tree, as each node is processed once.

How do I implement N-ary Tree Postorder Traversal iteratively?

Use DFS with a stack, process the children in reverse order, and ensure that nodes are only visited after all their children have been processed.

What’s the best way to manage state during N-ary tree traversal?

Use a stack to manage the nodes and keep track of the traversal order. Make sure children are visited before their parent node to maintain postorder sequence.

What if the tree is empty? How should I handle that in the traversal?

If the tree is empty, return an empty list, as there are no nodes to traverse.

terminal

Solution

Solution 1: Recursion

We can recursively traverse the entire tree. For each node, we first recursively call the function for each of the node's children, then add the node's value to the answer.

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"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""


class Solution:
    def postorder(self, root: 'Node') -> List[int]:
        def dfs(root):
            if root is None:
                return
            for child in root.children:
                dfs(child)
            ans.append(root.val)

        ans = []
        dfs(root)
        return ans

Solution 2: Iteration (Stack Implementation)

We can also solve this problem iteratively.

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"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""


class Solution:
    def postorder(self, root: 'Node') -> List[int]:
        def dfs(root):
            if root is None:
                return
            for child in root.children:
                dfs(child)
            ans.append(root.val)

        ans = []
        dfs(root)
        return ans

Continue Practicing

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