• Expected Output: 7
• Justification: The maximum difference is between node 1 and its ancestor 8, which is 7.

Solution

To solve this problem, we adopt a depth-first search (DFS) strategy, exploring each path from the root to the leaf nodes while keeping track of the maximum and minimum values encountered along each path. This approach is effective because it allows us to calculate the difference between the current node and its ancestors' minimum and maximum values in a single traversal, ensuring efficiency and simplicity.

The key is to update the minimum and maximum values as we move down the tree and calculate the difference at each node, comparing it with the global maximum difference found so far. This strategy is chosen for its ability to explore all possible ancestor-descendant pairs without redundant calculations or needing to store additional data structures that represent all ancestors of each node.

Step-by-step Algorithm

1. Start with the root node of the binary tree, passing the node's value as both the minimum and maximum values encountered so far to a recursive depth-first search (DFS) function.

2. DFS function:

   • If the current node is null, return the difference between the maximum and minimum values encountered so far on this path.
   • Update the minimum value if the current node's value is less than the current minimum.
   • Update the maximum value if the current node's value is greater than the current maximum.
   • Recursively call the DFS function for the left child of the current node, passing the updated minimum and maximum values.
   • Recursively call the DFS function for the right child of the current node, passing the updated minimum and maximum values.
   • Calculate the maximum difference found in the left and right subtree calls and return it.

3. Return the result of the DFS call started with the root node, which gives the maximum difference between any node and its ancestor in the binary tree.

Algorithm Walkthrough

Let's consider the input: root = [5, 3, 8, 0, null, 4, 7, 1].

Tree Structure:

The given tree can be visualized as follows:

        5
       / \
      3   8
     /   / \
    0   4   7
   /
  1

Step-by-Step Walkthrough:

1. Start at the Root (Node 5):

   • Current Node: 5
   • Current Min and Max: Both initialized to the value of the root, which is 5 (min = 5, max = 5).
   • Recursively call DFS on the left child (Node 3).

2. Move to the Left Child (Node 3):

   • Current Node: 3
   • Update Min and Max:
     • min = min(5, 3) = 3
     • max = max(5, 3) = 5
   • Recursively call DFS on the left child (Node 0).

3. Move to the Left Child (Node 0):

   • Current Node: 0
   • Update Min and Max:
     • min = min(3, 0) = 0
     • max = max(5, 0) = 5
   • Recursively call DFS on the left child (Node 1).

4. Move to the Left Child (Node 1):

   • Current Node: 1
   • Update Min and Max:
     • min = min(0, 1) = 0
     • max = max(5, 1) = 5
   • Both left and right children are null:
     • Return the difference max - min = 5 - 0 = 5.
   • Go back to the Node 0.

5. Node 0 - Right Child is Null:

   • Since the right child of Node 0 is null, the right difference is not calculated.
   • Return the maximum difference obtained from the left subtree: 5.
   • Go back to the Node 3.

6. Node 3 - Right Child is Null:

   • Since the right child of Node 3 is null, the right difference is not calculated.
   • Return the maximum difference obtained from the left subtree: 5.
   • Go back to the Node 5.

7. Move to the Right Child (Node 8):

   • Current Node: 8
   • Update Min and Max:
     • min = min(5, 8) = 5
     • max = max(5, 8) = 8
   • Recursively call DFS on the left child (Node 4).

8. Move to the Left Child (Node 4):

   • Current Node: 4
   • Update Min and Max:
     • min = min(5, 4) = 4
     • max = max(8, 4) = 8
   • Both left and right children are null:
     • Return the difference max - min = 8 - 4 = 4.
   • Go back to the Node 8.

9. Move to the Right Child (Node 7):

   • Current Node: 7
   • Update Min and Max:
     • min = min(5, 7) = 5
     • max = max(8, 7) = 8
   • Both left and right children are null:
     • Return the difference max - min = 8 - 5 = 3.
   • Go back to the Node 8.

10. Node 8 - Combine Left and Right Results:

    • Left difference: 4 (from Node 4)
    • Right difference: 3 (from Node 7)
    • Maximum difference for Node 8: max(4, 3) = 4
    • Go back to the Node 5.

11. Node 5 - Combine Left and Right Results:

    • Left difference: 5 (from Node 3's subtree)
    • Right difference: 4 (from Node 8's subtree)
    • Maximum difference for Node 5: max(5, 4) = 5

Final Output:

• The maximum difference between a node and its ancestor for the given tree [5, 3, 8, 0, null, 4, 7, 1] is 5.

Code