Binary Tree Postorder Traversal Leetcode Problem 145 [Python]
In the world of data structures and algorithms, binary trees are a fundamental concept.
These hierarchical structures play a crucial role in various computer science problems and are commonly encountered in coding interviews.
In this blog post, we'll delve into the Binary Tree Postorder Traversal problem, which is categorized as an "Easy" problem on LeetCode and falls under the "Trees" category.
We'll provide a Python solution for this problem, along with a detailed explanation of our approach.
Problem Overview
The Binary Tree Postorder Traversal problem on LeetCode is described as follows:
Problem Statement:
Given the root of a binary tree, you need to implement the "postorder" traversal of its nodes' values.
In a postorder traversal, you visit the left subtree, then the right subtree, and finally the root node.
Here's a visual representation of a binary tree:
1
/
2 3
/
4 5
A postorder traversal of this tree would yield the following sequence: [2, 4, 5, 3, 1].
Now, let's dive into the details of our solution.
Understanding Binary Tree Postorder Traversal Constraints
Before we jump into the coding part, it's essential to understand the problem constraints.
Here are the constraints for this problem:
- The number of nodes in the tree falls within the range
[0, 100]. - The values of the nodes are in the range
[-100, 100].
These constraints ensure that we are dealing with a reasonably sized binary tree, which allows us to use an efficient approach to solve the problem.
Binary Tree Postorder Traversal LeetCode Problem Solution
1. Bruteforce Approach
The problem of binary tree postorder traversal can be approached using both recursive and iterative methods.
We will begin by presenting an iterative solution using a stack, which is often more challenging than the recursive approach.
Let's start by initializing a stack and a list to store the final result:
def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
stack = [root]
visit = [False]
res = []
We maintain two stacks here.
The stack stores nodes, and the visit array helps us keep track of whether a node has been visited.
Now, let's dive into the main logic:
while stack:
cur, v = stack.pop(), visit.pop()
if cur:
if v:
res.append(cur.val)
else:
stack.append(cur)
visit.append(True)
stack.append(cur.right)
visit.append(False)
stack.append(cur.left)
visit.append(False)
In the while loop, we pop elements from the stack and their corresponding visit status.
If cur is not None (i.e., there's a valid node), we check if it has been visited.
If it has been visited, we add its value to the result list res.
If it hasn't been visited, we push it back onto the stack with visit status set to True and push its right and left children onto the stack with visit status set to False.
This iterative solution simulates the recursive postorder traversal by explicitly managing the traversal stack.
It ensures that the values are appended to the result list in the correct order.
2. An Efficient Approach with Stack
The previous approach, while being correct, uses two separate stacks to keep track of nodes and their visit status.
In this more efficient approach, we'll use a single stack to achieve the same result.
Here's the code for this improved approach:
def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
stack = [(root, False)]
res = []
while stack:
cur, visited = stack.pop()
if cur is None:
continue
if visited:
res.append(cur.val)
else:
stack.append((cur, True))
stack.append((cur.right, False))
stack.append((cur.left, False))
return res
In this approach, we use a tuple (node, visited) to represent each node in the stack.
The visited flag indicates whether the node has been visited.
We start by pushing the root node onto the stack with visited set to False.
In the while loop, we pop nodes from the stack, and if the node is None, we continue to the next iteration.
If the node has already been visited, we append its value to the result list.
If it hasn't been visited, we push it back onto the stack with visited set to True and push its right and left children with visited set to False.
This improved approach uses a single stack and simplifies the logic while achieving the same result.
It is a more efficient and elegant solution for the Binary Tree Postorder Traversal problem.
Binary Tree Postorder Traversal Python Code Solution
Now that we've explained both the brute-force and efficient approaches to solve the Binary Tree Postorder Traversal problem, let's provide the Python code for the efficient approach:
def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
stack = [(root, False)]
res = []
while stack:
cur, visited = stack.pop()
if cur is None:
continue
if visited:
res.append(cur.val)
else:
stack.append((cur, True))
stack.append((cur.right, False))
stack.append((cur.left, False))
return res
This code defines a Python function postorderTraversal that takes the root node of a binary tree as input and returns the postorder traversal of its nodes' values as a list.
Time and Space Complexity
It's important to analyze the time and space complexity of our solution to understand its efficiency.
Time Complexity:
The time complexity of our efficient solution is O(N), where N is the number of nodes in the binary tree.
We visit each node exactly once, and the stack operations take constant time.
Space Complexity:
The space complexity is also O(N) in the worst case, which occurs when the binary tree is skewed, and all nodes are pushed onto the stack.
We use a stack to store the nodes, and the size of the stack can be at most N.
Reasoning Behind Our Approach
The Binary Tree Postorder Traversal problem is a classic example of traversing a binary tree while maintaining the order of visiting the nodes.
We discussed two approaches, the iterative solution with two stacks and the more efficient solution with a single stack.
In the iterative solution, we used two stacks to simulate the recursive postorder traversal.
By explicitly managing the traversal stack and keeping track of visited nodes, we achieved the desired order of values in the result list.
The efficient approach introduced a more elegant way to solve the problem using a single stack.
We used a tuple to represent each node and its visit status, simplifying the logic while maintaining efficiency.
The reasoning behind these approaches lies in the understanding of postorder traversal, where we explore the left and right subtrees before visiting the root.
By mimicking this process using a stack, we
successfully solved the problem.
Related Interview Questions By Company:
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Conclusion
In this blog post, we tackled the Binary Tree Postorder Traversal problem, a common coding interview question.
We provided a Python solution that employs an efficient approach with a single stack, achieving the desired postorder traversal of a binary tree.
We discussed the problem constraints, the two main approaches to solving it, and the reasoning behind our efficient solution.
Understanding and mastering binary tree traversal is essential for any aspiring programmer or software engineer, and this problem serves as a great exercise in that regard.
We hope this guide has been helpful in your journey to mastering data structures and algorithms.
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Question Link: Binary Tree Postorder Traversal on LeetCode
Remember, practice makes perfect, and happy coding!
