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Trim Dead Nodes from Binary Tree

You are given the reference to the root of a binary tree and are asked to trim the tree of "dead" nodes. A dead node is a node whose value is listed in the provided dead array. Once the tree has been trimmed of all dead nodes, return a list containing references to the roots of all the remaining segments of the tree.

Note: When a dead node is removed, its children become roots of new segments (if they are not dead themselves).

Example(s)​

Example 1:

Input:
      3
    /   \
   1     7
 /  \   /  \
2    8 4    6

dead = [7, 8]

Output: [3, 4, 6]

Explanation:
- Node 7 is dead, so it's removed. Its children (4 and 6) become new roots.
- Node 8 is dead, so it's removed. It has no children.
- The original root 3 remains (it's not dead).
- Result: [3, 4, 6]

Example 2:

Input:
      5
    /   \
   3     2
  / \   / \
 1   4 6   7

dead = [3, 2]

Output: [5, 1, 4, 6, 7]

Explanation:
- Node 3 is dead, so its children (1 and 4) become new roots.
- Node 2 is dead, so its children (6 and 7) become new roots.
- The original root 5 remains.
- Result: [5, 1, 4, 6, 7]

Example 3:

Input:
      1
    /   \
   2     3

dead = [1]

Output: [2, 3]

Explanation:
- Root 1 is dead, so it's removed.
- Its children (2 and 3) become new roots.
- Result: [2, 3]

Solution​

The solution uses post-order traversal with a result collector:

  1. Post-order traversal: Process children before parent (left → right → root)
  2. Check if node is dead: If dead, add its non-dead children to result and return null
  3. Recursively process children: Update children by removing dead nodes
  4. Collect roots: When a dead node is found, its children become new roots
  5. Handle original root: If original root is not dead, add it to result

JavaScript Solution

/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/

/**
* Trim dead nodes from binary tree and return roots of remaining segments
* @param {TreeNode} root - Root of the binary tree
* @param {number[]} dead - Array of dead node values
* @return {TreeNode[]} - Array of root references for remaining segments
*/
function trimDeadNodes(root, dead) {
const deadSet = new Set(dead); // For O(1) lookup
const result = [];

/**
 * Recursively process the tree and collect roots
 * @param {TreeNode} node - Current node
 * @return {TreeNode|null} - Modified node or null if dead
 */
function dfs(node) {
  if (!node) {
    return null;
  }
  
  // Check if current node is dead
  if (deadSet.has(node.val)) {
    // Node is dead, so its children become potential new roots
    // Process children first (they might also be dead)
    const leftChild = dfs(node.left);
    const rightChild = dfs(node.right);
    
    // Add non-dead children to result as new roots
    if (leftChild && !deadSet.has(leftChild.val)) {
      result.push(leftChild);
    }
    if (rightChild && !deadSet.has(rightChild.val)) {
      result.push(rightChild);
    }
    
    // Return null to remove this dead node
    return null;
  }
  
  // Node is not dead, recursively process children
  node.left = dfs(node.left);
  node.right = dfs(node.right);
  
  return node;
}

// Process the tree
const processedRoot = dfs(root);

// If original root is not dead, add it to result
if (processedRoot && !deadSet.has(processedRoot.val)) {
  result.push(processedRoot);
}

return result;
}

// Helper function to create a tree node
function TreeNode(val, left, right) {
this.val = (val===undefined ? 0 : val);
this.left = (left===undefined ? null : left);
this.right = (right===undefined ? null : right);
}

// Helper function to print tree values
function getValues(nodes) {
return nodes.map(node => node.val);
}

// Test case 1: [3, 1, 7, 2, 8, 4, 6], dead = [7, 8]
const tree1 = new TreeNode(3);
tree1.left = new TreeNode(1);
tree1.right = new TreeNode(7);
tree1.left.left = new TreeNode(2);
tree1.left.right = new TreeNode(8);
tree1.right.left = new TreeNode(4);
tree1.right.right = new TreeNode(6);
const result1 = trimDeadNodes(tree1, [7, 8]);
console.log('Example 1:', getValues(result1)); // [3, 4, 6]

// Test case 2: [5, 3, 2, 1, 4, 6, 7], dead = [3, 2]
const tree2 = new TreeNode(5);
tree2.left = new TreeNode(3);
tree2.right = new TreeNode(2);
tree2.left.left = new TreeNode(1);
tree2.left.right = new TreeNode(4);
tree2.right.left = new TreeNode(6);
tree2.right.right = new TreeNode(7);
const result2 = trimDeadNodes(tree2, [3, 2]);
console.log('Example 2:', getValues(result2)); // [5, 1, 4, 6, 7]

// Test case 3: [1, 2, 3], dead = [1]
const tree3 = new TreeNode(1);
tree3.left = new TreeNode(2);
tree3.right = new TreeNode(3);
const result3 = trimDeadNodes(tree3, [1]);
console.log('Example 3:', getValues(result3)); // [2, 3]
Output:
Click "Run Code" to execute the code and see the results.

Alternative Solution (Two-Pass)​

Here's an alternative approach that first collects all roots, then trims:

/**
 * Alternative: Two-pass approach
 * First pass: identify and collect roots
 * Second pass: trim the tree
 */
function trimDeadNodesTwoPass(root, dead) {
  const deadSet = new Set(dead);
  const roots = [];
  
  // First pass: collect all roots (original + children of dead nodes)
  function collectRoots(node, isRoot) {
    if (!node) return;
    
    if (isRoot && !deadSet.has(node.val)) {
      roots.push(node);
    }
    
    const isDead = deadSet.has(node.val);
    collectRoots(node.left, isDead);
    collectRoots(node.right, isDead);
  }
  
  collectRoots(root, true);
  
  // Second pass: trim dead nodes
  function trim(node) {
    if (!node) return null;
    if (deadSet.has(node.val)) return null;
    
    node.left = trim(node.left);
    node.right = trim(node.right);
    return node;
  }
  
  // Trim each root
  return roots.map(trim).filter(Boolean);
}

Complexity​

  • Time Complexity: O(n) - Where n is the number of nodes. We visit each node exactly once.
  • Space Complexity: O(h + m) - Where h is the height of the tree (recursion stack) and m is the number of dead nodes (for the set). In the worst case, O(n) for skewed tree.

Approach​

The solution uses post-order traversal with root collection:

  1. Convert dead array to set: For O(1) lookup time
  2. Post-order traversal: Process children before parent
  3. Dead node handling:
    • When a dead node is found, recursively process its children
    • Add non-dead children to the result list (they become new roots)
    • Return null to remove the dead node
  4. Non-dead node handling:
    • Recursively process and update children
    • Return the node (possibly with modified children)
  5. Original root handling: After processing, if the original root is not dead, add it to result

Key Insights​

  • Post-order traversal ensures children are processed before parent
  • Children of dead nodes become roots: This is the key insight - when a node is removed, its children become independent trees
  • Set for dead values: O(1) lookup instead of O(m) array search
  • In-place modification: We modify the tree structure in place
  • Root collection: We collect roots as we find dead nodes, not at the end

Step-by-Step Example​

Let's trace through Example 1: Tree with nodes [3, 1, 7, 2, 8, 4, 6], dead = [7, 8]

Initial tree:
      3
    /   \
   1     7
 /  \   /  \
2    8 4    6

Post-order traversal: 2 → 8 → 1 → 4 → 6 → 7 → 3

Step 1: Process node 2 (leaf, not dead)
  - Return node 2

Step 2: Process node 8 (leaf, dead)
  - Add children to result: none (leaf node)
  - Return null

Step 3: Process node 1 (not dead)
  - left = 2, right = null (8 was removed)
  - Return node 1

Step 4: Process node 4 (leaf, not dead)
  - Return node 4

Step 5: Process node 6 (leaf, not dead)
  - Return node 6

Step 6: Process node 7 (dead)
  - left = 4, right = 6 (both not dead)
  - Add 4 and 6 to result
  - Return null

Step 7: Process node 3 (root, not dead)
  - left = 1, right = null (7 was removed)
  - Return node 3

Final: Add root 3 to result
Result: [3, 4, 6]

Edge Cases​

  • Root is dead: Children become new roots
  • All nodes are dead: Return empty array
  • No dead nodes: Return array with original root
  • Dead node with dead children: Only non-dead grandchildren become roots
  • Single node tree: If dead, return empty; if not, return [root]
  • Empty tree: Return empty array

Important Notes​

  • Node references: The problem asks for references to nodes, not values
  • Tree modification: The original tree structure is modified (dead nodes removed)
  • Multiple roots: Result can contain multiple root references
  • Children become roots: When a dead node is removed, its children become roots of new segments

Takeaways​

  • Post-order traversal is useful when children need to be processed before parent
  • Root collection during traversal is more efficient than collecting at the end
  • Set for lookups improves time complexity from O(nm) to O(n)
  • In-place modification is space-efficient
  • Understanding the problem - children of dead nodes become roots - is crucial