Report Tree Levels and Balanced Subtrees

Question

The question asks the candidate to implement a solution that reports the levels of nodes in a binary tree and identifies whether each subtree is balanced.

Codemia · Accepted Answer

To solve the problem, I will use a depth-first search (DFS) to traverse the binary tree. During the traversal, I will keep track of the level of each node and check if the subtree rooted at each node is balanced. A subtree is considered balanced if the height difference between its left and right subtrees is at most one.

The algorithm will involve a recursive function that processes each node. The function will return the height of the subtree and whether it is balanced. Here is a Python implementation:

```python
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def check_balanced_and_report_levels(node, level=0):
    if not node:
        return 0, True
    left_height, left_balanced = check_balanced_and_report_levels(node.left, level + 1)
    right_height, right_balanced = check_balanced_and_report_levels(node.right, level + 1)
    is_balanced = abs(left_height - right_height) <= 1
    print(f'Node value: {node.val}, Level: {level}, Is Balanced: {is_balanced}')
    return max(left_height, right_height) + 1, is_balanced
```
This function prints the value of each node, its level, and whether the subtree is balanced.

The time complexity of this algorithm is O(n), where n is the number of nodes in the tree, as we visit each node once. The space complexity is O(h), where h is the height of the tree, due to the recursive call stack.

A balanced tree ensures that the height of the tree remains logarithmic relative to the number of nodes, which optimizes search, insert, and delete operations. In this implementation, we check the balance condition during the DFS traversal, which allows us to efficiently determine the balance status of each subtree.

Report Tree Levels and Balanced Subtrees

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