medium Sum of Nodes with Even-Valued Grandparent

Problem Statement

Given the root of a binary tree, return the total sum of the values of all nodes that have a grandparent with an even number. If no such nodes exist, return 0.

A grandparent of a node is defined as the parent of its parent, if both exist.

Examples

Example 1:

• Input: root = [4, 2, 6, 3, 5, null, 8]
• Expected Output: 16

• Justification: Node values 3, 5, and 8 have 4 as their grandparent. Since 4 is even, their sum is 3 + 5 + 8 = 16.

Example 2:

• Input: root = [2, 3, 8, null, 5, 7, null, 2, 3, 4, 5]
• Expected Output: 21

• Justification: Node value 5 and 7 has 2 as its grandparent, and Node value 4 and 5 has grandparent 8. So, sum = 5 + 7 + 4 + 5 = 21.

Example 3:

• Input: root = [5, 9, 12, null, null, 8, 4]
• Expected Output: 0

• Justification: Node values 8 and 4 has 5 as its grandparent. However, since 5 is not even, they are not included. So, the answer is 0.

Constraints:

• The number of nodes in the tree is in the range [1, 10⁴].
• 1 <= Node.val <= 100

Solution

The algorithm for finding the sum of nodes with even-valued grandparents begins by checking if the root is null, returning 0 if true. We then initialize a queue for BFS traversal, starting with the root node, and a variable sum to keep track of the result. For each node, if its value is even, we check its left and right children for grandchildren, adding their values to the sum if they exist. All non-null children are added to the queue to continue the BFS traversal. This process repeats until all nodes are processed, and the final sum is returned.

Step-by-Step Algorithm

Step 1: Check if the tree is empty:

• If the root is null, return 0. This handles the case where the binary tree is empty.

Step 2: Initialize the sum variable:

• A variable sum is initialized to 0. This will keep track of the total sum of the node values that meet the condition of having an even-valued grandparent.

Step 3: Initialize a queue for level order traversal (BFS):

• A Queue is used to perform BFS traversal starting from the root. The root node is added to the queue.

Step 4: Perform BFS traversal:

• While the queue is not empty, perform the following steps for each node:

  1. Step 4.1: Remove the front node from the queue:

     • Use queue.poll() to remove the current node for processing.

  2. Step 4.2: Check if the current node's value is even:

     • If current.val % 2 == 0, it means the current node has an even value.
     • If the current node is even-valued, we will check for its grandchildren.

  3. Step 4.3: Add the values of the grandchildren (if they exist):

     • Left grandchild check:
       • If the left child of the current node exists (current.left != null), then check for its children (i.e., grandchildren of the current node).
       • If the left grandchild exists (current.left.left != null), add its value to sum.
       • If the right grandchild exists (current.left.right != null), add its value to sum.
     • Right grandchild check:
       • If the right child of the current node exists (current.right != null), then check for its children.
       • If the left grandchild exists (current.right.left != null), add its value to sum.
       • If the right grandchild exists (current.right.right != null), add its value to sum.

  4. Step 4.4: Add the children of the current node to the queue for further processing:

     • If the left child exists (current.left != null), add it to the queue.
     • If the right child exists (current.right != null), add it to the queue.

Step 5: Return the total sum:

• After processing all nodes in the tree, return the value of sum, which represents the sum of all nodes with even-valued grandparents.

Algorithm Walkthrough

Step 1: Processing the Root Node (Value 4)

• We begin at the root node with the value 4. The queue initially contains only the root node: Queue = [4].

• The variable sum is initialized to 0.

• Since 4 is an even-valued node, we now check the values of its grandchildren:

  • Exploring children of the left child (Node 2):
    • The left child of 4 is 2, which has two children: 3 and 5. Both are the grandchildren of node 4.
    • We add the values of both 3 and 5 to sum.
    • Updated sum = 0 + 3 + 5 = 8.
  • Exploring children of the right child (Node 6):
    • The right child of 4 is 6. It has one child: 8, which is also a grandchild of node 4.
    • We add the value of 8 to sum.
    • Updated sum = 8 + 8 = 16.

• After processing the grandchildren of 4, we add its left child (2) and right child (6) to the queue for further processing.

  • Queue = [2, 6].

Step 2: Processing Node 2

• The next node in the queue is 2. We remove it from the queue: Queue = [6].

• Since 2 is an even-valued node, we now check the values of its grandchildren:

  • Exploring children of the left child (Node 3):
    • The left child of 2 is 3. Node 3 has no children, so there are no grandchildren to add to sum.
  • Exploring children of the right child (Node 5):
    • The right child of 2 is 5. Node 5 also has no children, so there are no grandchildren to add to sum.

• Since there are no grandchildren for node 2, the sum remains unchanged at 16.

• We then add its children (3 and 5) to the queue for further processing.

  • Queue = [6, 3, 5].

Step 3: Processing Node 6

• The next node in the queue is 6. We remove it from the queue: Queue = [3, 5].

• Since 6 is an even-valued node, we now check the values of its grandchildren:

  • Exploring children of the right child (Node 8):
    • The right child of 6 is 8. Node 8 has no children, so there are no grandchildren to add to sum.

• Since there are no grandchildren for node 6 that haven’t already been processed, the sum remains unchanged at 16.

• We then add its right child (8) to the queue for further processing.

  • Queue = [3, 5, 8].

Step 4: Processing Node 3

• The next node in the queue is 3. We remove it from the queue: Queue = [5, 8].

• Since 3 is an odd-valued node, we do not process its grandchildren.

• Node 3 has no children, so no further nodes are added to the queue.

Step 5: Processing Node 5

• The next node in the queue is 5. We remove it from the queue: Queue = [8].

• Since 5 is an odd-valued node, we do not process its grandchildren.

• Node 5 has no children, so no further nodes are added to the queue.

Step 6: Processing Node 8

• The last node in the queue is 8. We remove it from the queue: Queue = [].

• Since 8 is an even-valued node, we would normally check its grandchildren. However, node 8 has no children, so no further nodes are added to the queue.

Final Step: End of BFS Traversal

• The queue is now empty, so the BFS traversal is complete.
• The total sum of all grandchildren of even-valued nodes is 16.

Code

// class TreeNode {
//     int val;
//     TreeNode left;
//     TreeNode right;

//     // Constructor for initializing the node
//     TreeNode(int val) {
//         this.val = val;
//     }
// }

public class Solution {

  public int sumEvenGrandparent(TreeNode root) {
    if (root == null) return 0;

    int sum = 0; // Initialize sum to 0
    Queue<TreeNode> queue = new LinkedList<>(); // Queue for level order traversal
    queue.add(root); // Start with the root node

    while (!queue.isEmpty()) {
      TreeNode current = queue.poll(); // Get the next node from the queue

      // Check if current node has an even value
      if (current.val % 2 == 0) {
        // If so, add the values of all grandchildren
        if (current.left != null) {
          if (current.left.left != null) sum += current.left.left.val; // Add left grandchild
          if (current.left.right != null) sum += current.left.right.val; // Add right grandchild
        }
        if (current.right != null) {
          if (current.right.left != null) sum += current.right.left.val; // Add left grandchild
          if (current.right.right != null) sum += current.right.right.val; // Add right grandchild
        }
      }

      // Continue with the next level nodes
      if (current.left != null) queue.add(current.left);
      if (current.right != null) queue.add(current.right);
    }

    return sum;
  }

  // Main method to test the function
  public static void main(String[] args) {
    // Construct the first tree: [4, 2, 6, 3, 5, null, 8]
    TreeNode root1 = new TreeNode(4); // Root node
    root1.left = new TreeNode(2); // Left child of root
    root1.right = new TreeNode(6); // Right child of root
    root1.left.left = new TreeNode(3); // Left child of node 2
    root1.left.right = new TreeNode(5); // Right child of node 2
    root1.right.right = new TreeNode(8); // Right child of node 6

    // Construct the second tree: [2, 3, 8, null, 5, 7, null, 2, 3, 4, 5]
    TreeNode root2 = new TreeNode(2); // Root node
    root2.left = new TreeNode(3); // Left child of root
    root2.right = new TreeNode(8); // Right child of root
    root2.left.right = new TreeNode(5); // Right child of node 3
    root2.right.left = new TreeNode(7); // Left child of node 8
    root2.left.right.left = new TreeNode(2);
    root2.left.right.right = new TreeNode(3);
    root2.right.left.left = new TreeNode(4);
    root2.right.left.right = new TreeNode(5);

    // Create a Solution object to use the method
    Solution solution = new Solution();

    // Test the function with the first example
    System.out.println("Sum for tree 1: " + solution.sumEvenGrandparent(root1)); // Expected output: 16

    // Test the function with the second example
    System.out.println("Sum for tree 2: " + solution.sumEvenGrandparent(root2)); // Expected output: 21
  }
}

Complexity Analysis

Time Complexity

The time complexity of the algorithm is , where n is the number of nodes in the binary tree. This is because we traverse each node exactly once using a level order traversal, checking each node to see if it has an even-valued grandparent and adding the values of its grandchildren if applicable.

Space Complexity

The space complexity is , where m is the maximum number of nodes at any level of the binary tree. In the worst case, this can be proportional to the width of the tree. If the tree is a complete binary tree, this will be . This space is needed for the queue used in the level order traversal.