Given a binary tree, return the postorder traversal of its nodes' values.
Example:
Input:[1,null,2,3]
1 \ 2 / 3 Output:[3,2,1]
Follow up: Recursive solution is trivial, could you do it iteratively?
The core challenge of this problem is to traverse a binary tree in postorder, which means visiting the left subtree, then the right subtree, and finally the root node. This traversal is significant in various applications such as expression tree evaluations and syntax tree traversals in compilers.
Potential pitfalls include misunderstanding the traversal order and handling null nodes incorrectly.
To solve this problem, we can consider both recursive and iterative approaches:
The recursive approach is straightforward. We define a function that recursively visits the left subtree, then the right subtree, and finally appends the root node's value to the result list.
The iterative approach is more complex but can be achieved using a stack. We use two stacks to simulate the postorder traversal. The first stack is used to traverse the tree, and the second stack is used to store the nodes in postorder.
def postorder_traversal_recursive(root):
def helper(node, result):
if node:
helper(node.left, result) # Traverse left subtree
helper(node.right, result) # Traverse right subtree
result.append(node.val) # Visit root node
result = []
helper(root, result)
return result
def postorder_traversal_iterative(root):
if not root:
return []
stack1, stack2 = [root], []
result = []
while stack1:
node = stack1.pop()
stack2.append(node)
if node.left:
stack1.append(node.left) # Push left child to stack1
if node.right:
stack1.append(node.right) # Push right child to stack1
while stack2:
node = stack2.pop()
result.append(node.val) # Collect nodes in postorder
return result
Both the recursive and iterative solutions have a time complexity of O(n), where n is the number of nodes in the tree. This is because each node is visited exactly once.
The space complexity for the recursive solution is O(h), where h is the height of the tree, due to the call stack. For the iterative solution, the space complexity is O(n) due to the use of two stacks.
Potential edge cases include:
To test the solution comprehensively, consider the following test cases:
[]
[1]
[1, 2, null, 3]
[1, null, 2, null, 3]
[1, 2, 3, 4, 5, 6, 7]
When approaching tree traversal problems, it is helpful to:
Postorder tree traversal is a fundamental problem in computer science with various applications. Understanding both recursive and iterative approaches is crucial for solving more complex tree-related problems. Practice and familiarity with different traversal techniques will improve your problem-solving skills.
For further reading and practice, consider the following resources: