Solution Explanation for Even Odd Tree

This problem asks us to determine if a binary tree is an "Even-Odd Tree". An Even-Odd Tree satisfies these conditions:

1. Level Indexing: The root is at level 0, its children at level 1, and so on.
2. Even Levels: Nodes at even levels must have odd values and be strictly increasing from left to right.
3. Odd Levels: Nodes at odd levels must have even values and be strictly decreasing from left to right.

Both Breadth-First Search (BFS) and Depth-First Search (DFS) approaches are efficient solutions. Let's examine each:

Solution 1: Breadth-First Search (BFS)

BFS is well-suited for level-order traversal. We process nodes level by level, checking the conditions for each level.

Algorithm:

1. Initialization: Use a queue to store nodes for BFS traversal. Initialize even flag to track even/odd level.
2. Level Processing: Iterate while the queue is not empty. Process all nodes at the current level.
3. Even Level Check: If even is true (even level):
   • Check if the node's value is even or if it's not strictly increasing compared to the previous node. If either is true, the tree is not Even-Odd.
4. Odd Level Check: If even is false (odd level):
   • Check if the node's value is odd or if it's not strictly decreasing compared to the previous node. If either is true, the tree is not Even-Odd.
5. Update: After processing a level, toggle the even flag and move to the next level.

Time Complexity: O(N), where N is the number of nodes. We visit each node once. Space Complexity: O(W), where W is the maximum width of the tree. In the worst case (a complete binary tree), W can be N, so space complexity is O(N).

Solution 2: Depth-First Search (DFS)

DFS uses recursion to traverse the tree. We maintain the current level and check the conditions recursively.

Algorithm:

1. Helper Function (dfs): A recursive helper function takes the current node and level as input.
2. Base Case: If the node is null, return true (it satisfies the condition).
3. Even/Odd Check: Similar to BFS, check the conditions based on the level's parity.
4. Update: If all checks pass, recursively call dfs on the left and right children, updating the level.
5. Hash Table: A hash table (d in the code) stores the last seen value for each level to efficiently compare with the next node's value within the level.

Time Complexity: O(N). We visit each node once. Space Complexity: O(H), where H is the height of the tree. In the worst-case scenario (a skewed tree), H can be N, leading to O(N) space complexity.

Code Examples (Python):

BFS (Python):

from collections import deque
 
def isEvenOddTree(root):
    q = deque([root])
    even = True
    while q:
        level_size = len(q)
        prev = (0 if even else float('inf')) #Initialize prev appropriately for even/odd
        for _ in range(level_size):
            node = q.popleft()
            if even:
                if node.val % 2 == 0 or node.val <= prev:
                    return False
            else:
                if node.val % 2 != 0 or node.val >= prev:
                    return False
            prev = node.val
            if node.left:
                q.append(node.left)
            if node.right:
                q.append(node.right)
        even = not even
    return True
 

DFS (Python):

def isEvenOddTree(root):
    d = {}  # Hash table to store last seen value for each level
 
    def dfs(node, level):
        if not node:
            return True
        even = level % 2 == 0
        prev = d.get(level, 0 if even else float('inf')) #Handle default for even/odd
        if even:
            if node.val % 2 == 0 or node.val <= prev:
                return False
        else:
            if node.val % 2 != 0 or node.val >= prev:
                return False
        d[level] = node.val
        return dfs(node.left, level + 1) and dfs(node.right, level + 1)
 
    return dfs(root, 0)
 

Both solutions are valid and have similar time and space complexities. The choice between BFS and DFS might depend on personal preference or specific constraints of a problem. BFS might be slightly more intuitive for level-order processing, while DFS might be easier to implement recursively.