Leetcode 1080: Insufficient Nodes in Root to Leaf Paths
You are given the root of a binary tree and an integer limit. Your task is to delete all nodes in the tree that are considered insufficient. A node is insufficient if every root-to-leaf path passing through that node has a sum strictly less than the given limit. A leaf is defined as a node with no children. Return the root of the resulting binary tree after the deletions.
Problem
Approach
Steps
Complexity
Input: The input consists of the root of a binary tree represented by its root node and an integer limit.
Example: Input: root = [2, 3, 4, 5, -10, -10, 8, 10], limit = 15
Constraints:
• 1 <= Number of nodes in the tree <= 5000
• -10^5 <= Node value <= 10^5
• -10^9 <= limit <= 10^9
Output: The output should be the root node of the resulting binary tree after deleting insufficient nodes.
Example: Output: [2, 3, 4, 5, null, null, 10]
Constraints:
• The structure of the binary tree should be preserved after deleting the insufficient nodes.
Goal: The goal is to traverse the tree and prune any nodes whose root-to-leaf path does not satisfy the sum condition with respect to the limit.
Steps:
• 1. Perform a post-order traversal of the tree.
• 2. At each node, compute the sum of the path from root to leaf, considering all descendants.
• 3. If the sum at the current node is less than the limit, delete the node.
• 4. Return the modified tree after pruning the insufficient nodes.
Goal: The solution must handle trees with varying numbers of nodes efficiently.
Steps:
• 1 <= Number of nodes in the tree <= 5000
• -10^5 <= Node value <= 10^5
• -10^9 <= limit <= 10^9
Assumptions:
• The input tree is a valid binary tree.
• The limit value is a valid integer within the specified range.
• Input: Input: root = [3, 1, 2, 4, 5, -99, 6], limit = 10
• Explanation: In this case, the path from root to leaf through nodes 3 -> 1 -> 4 has a sum of 8, which is less than the limit. Thus, node 1 and its descendants (4) will be deleted. The resulting tree will have root 3, with the right child 2, and the leaf nodes 6 and 5 remaining.
• Input: Input: root = [2, 3, 4, 5, 6, 7, 8], limit = 18
• Explanation: The tree contains multiple paths where the sum exceeds the limit. After pruning, only the nodes forming valid paths above the limit are retained.
Approach: To solve this problem, a post-order depth-first search (DFS) approach can be employed, where each node is processed after its children, and insufficient nodes are removed based on the sum condition.
Observations:
• A bottom-up DFS traversal works well because we need to decide whether to prune a node after considering its children.
• Using recursion allows us to handle the pruning and return the modified tree efficiently.
Steps:
• 1. Start by performing a post-order DFS on the tree.
• 2. For each node, check if its left and right children need to be pruned (i.e., if they lead to an insufficient path).
• 3. If both children are insufficient, delete the node.
• 4. After pruning, return the updated tree structure.
Empty Inputs:
• If the tree is empty (i.e., root is null), return null.
Large Inputs:
• The solution must handle large trees (up to 5000 nodes) efficiently.
Special Values:
• If the limit is extremely large or small, the algorithm should still work within the given constraints.
Constraints:
• The solution should efficiently prune insufficient nodes in large binary trees.
TreeNode* sufficientSubset(TreeNode* root, int limit) {
if(!root) return NULL;
if(root->left == NULL && root->right == NULL)
return root->val < limit ? NULL : root;
root->left = sufficientSubset(root->left, limit - root->val);
root->right= sufficientSubset(root->right, limit - root->val);
return root->left == root->right ? NULL : root;
}
1 : Function Definition
TreeNode* sufficientSubset(TreeNode* root, int limit) {
This line defines the function `sufficientSubset`, which takes a pointer to the root of a binary tree and an integer `limit`. It returns the root of the modified tree after removing insufficient nodes.
2 : Base Case - Empty Node
if(!root) return NULL;
This checks if the current node is `NULL`. If the node is empty, it returns `NULL`, signaling the end of this path in the tree.
3 : Base Case - Leaf Node
if(root->left == NULL && root->right == NULL)
This checks if the current node is a leaf (i.e., has no left or right child).
4 : Leaf Node Value Check
return root->val < limit ? NULL : root;
If the node is a leaf, it checks whether its value is less than the `limit`. If it is, the node is removed (returns `NULL`), otherwise, it is kept.
5 : Recursive Call - Left Child
root->left = sufficientSubset(root->left, limit - root->val);
This recursively calls the `sufficientSubset` function on the left child of the current node, reducing the `limit` by the current node's value.
6 : Recursive Call - Right Child
root->right= sufficientSubset(root->right, limit - root->val);
This recursively calls the `sufficientSubset` function on the right child of the current node, similarly adjusting the `limit`.
7 : Return Condition
return root->left == root->right ? NULL : root;
This checks if both the left and right children are `NULL` (meaning both children were removed). If so, the current node is also removed (returns `NULL`), otherwise, it is returned.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: In the worst case, we visit all nodes in the tree once, leading to a linear time complexity with respect to the number of nodes.
Best Case: O(1)
Worst Case: O(h)
Description: The space complexity is proportional to the height of the tree due to the recursion stack in the DFS traversal.
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