Unit 9 Session 1 (Click for link to problem statements)
Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
HAPPY CASE
1 (root)
/ \
2 3
/ \ /
4 5 6
Input: root
Output: True
Explanation: Removing the edge between nodes 1 and 2 can split the tree into two trees with 3 nodes each.
EDGE CASE
1 (root)
Input: root
Output: False
Explanation: Only one node cannot be split into equal parts.
Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.
For Tree problems, we want to consider the following approaches:
Plan the solution with appropriate visualizations and pseudocode.
General Idea: First, count the total number of nodes in the tree. Then, perform another DFS to check if any subtree has exactly half the total number of nodes. If such a subtree is found, the tree can be split into two equal parts by removing the edge leading to this subtree.
1) Define a helper function `count_nodes(root)` to count the total number of nodes in the tree.
a) If the current node is None, return 0.
b) Recursively count nodes in the left and right subtrees and add 1 for the current node.
2) Define another helper function `can_split_helper(root, total_nodes)` to check if any subtree can form half the total number of nodes.
a) If the current node is None, return 0.
b) Recursively count nodes in the left and right subtrees.
c) If any subtree count equals half the total number of nodes, set a flag to True.
d) Return the total count of nodes for the current subtree.
3) In the main function `can_split(root)`, calculate the total number of nodes.
4) If the total number of nodes is odd, return False.
5) Use the helper function to check if the tree can be split, and return the result.
⚠️ Common Mistakes
Implement the code to solve the algorithm.
class TreeNode:
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
def count_nodes(root):
if not root:
return 0
return 1 + count_nodes(root.left) + count_nodes(root.right)
def can_split_helper(root, total_nodes):
if not root:
return 0
left_count = can_split_helper(root.left, total_nodes)
right_count = can_split_helper(root.right, total_nodes)
# If any subtree has half of the total nodes, we can split the tree
if left_count == total_nodes // 2 or right_count == total_nodes // 2:
can_split_helper.found = True
return 1 + left_count + right_count
def can_split(root):
total_nodes = count_nodes(root)
if total_nodes % 2 != 0:
return False
can_split_helper.found = False
can_split_helper(root, total_nodes)
return can_split_helper.found
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Assume N
represents the number of nodes in the tree.
O(N)
because we need to traverse all the nodes in the tree.O(N)
for the recursion stack in the worst case when the tree is completely unbalanced.