Correct a Binary Tree

Problem Statement

In this problem, we are given a binary tree that contains a specific defect: there is one node (referred to as fromNode) whose right child incorrectly points to another node (toNode) that exists at the same depth but on its right. This scenario is anomalous as it defies the typical structure of a binary tree where each node can only have children directly beneath it.

The task is to modify this tree such that the subtree rooted at the defective fromNode, including all its descendants, is removed from the tree. However, the node to which it incorrectly points (toNode) should remain untouched, along with the rest of the tree. This ensures we correct the structure while preserving as much of the original tree as possible.

Examples

Example 1

Input:

root = [1,2,3], fromNode = 2, toNode = 3

Output:

[1,null,3]

Explanation:

The node with value 2 is invalid, so remove it.

Example 2

Input:

root = [8,3,1,7,null,9,4,2,null,null,null,5,6], fromNode = 7, toNode = 4

Output:

[8,3,1,null,null,9,4,null,null,5,6]

Explanation:

The node with value 7 is invalid, so remove it and the node underneath it, node 2.

Constraints

• The number of nodes in the tree is in the range [3, 104].
• -109 <= Node.val <= 109
• All Node.val are unique.
• fromNode != toNode
• fromNode and toNode will exist in the tree and will be on the same depth.
• toNode is to the right of fromNode.
• fromNode.right is null in the initial tree from the test data.

Approach and Intuition

1. First, we need to identify the invalid node (fromNode), which has an incorrect right child pointing to another node on the same level but to its right (toNode).

2. As we traverse the tree, we maintain a record of nodes visited at each depth using a set or dictionary. This will help in detecting when a node’s right child is already present in the set of that particular depth, indicating the existence of the incorrect linkage.

3. To detect the invalid node, during each step of the tree traversal (typically a breadth-first search (BFS) to ensure level-order access), check if the right child of the current node is a node that exists in our depth-tracking structure. If so, this node is our fromNode.

4. Once the invalid node is identified, modify the tree by removing this node and all nodes beneath it, while ensuring not to remove the node that was incorrectly pointed to (toNode).

5. The traversal and modification should be carefully implemented to ensure that the tree’s integrity and node relationships outside the defective structure are maintained.

By following the above steps, we correct the binary tree consistently according to the problem's requirements. This methodology leverages the unique constraints specified, such as the position relationship between fromNode and toNode and their presence at the same depth.

Solutions

class Solution {
public:
    TreeNode* modifyTree(TreeNode* root) {
        // BFS processing queue, keeps node along with its parent
        queue<pair<TreeNode*, TreeNode*>> bfsQueue;
        bfsQueue.push({root, nullptr});
    
        // Process tree level-by-level
        while (!bfsQueue.empty()) {
            // Get number of nodes at current level
            int levelSize = bfsQueue.size();
    
            // Current level nodes tracking
            unordered_set<TreeNode*> currentLevelNodes;
    
            // Handle each node at this level
            for (int idx = 0; idx < levelSize; ++idx) {
                // Retrieve node and its parent from queue
                auto [currentNode, parentNode] = bfsQueue.front();
                bfsQueue.pop();
    
                // Check if right child is already processed
                if (currentLevelNodes.count(currentNode->right)) {
                    // Correct tree by removing defective node
                    if (parentNode->left == currentNode) {
                        parentNode->left = nullptr;
                    } else {
                        parentNode->right = nullptr;
                    }
                    return root;
                }
    
                // Mark current node as visited
                currentLevelNodes.insert(currentNode);
    
                // Enqueue children for next level exploration
                if (currentNode->right) {
                    bfsQueue.push({currentNode->right, currentNode});
                }
                if (currentNode->left) {
                    bfsQueue.push({currentNode->left, currentNode});
                }
            }
        }
    
        // Return modified root
        return root;
    }
};