Count Good Nodes in Binary Tree
A node is good if from the root to that node, there is no node with a value greater than it.
Given a binary tree (level-order array), return the count of good nodes.
Example:
3,1,4,3,null,1,5
4
- The binary tree is constructed from the level-order array: the root node is 3, its children are 1 and 4, and so on.
- We start from the root node and traverse down, checking each node's value against the maximum value seen so far: the root node 3 is good, its child node 1 is good because 1<3, and the child node 4 is not good because 4>3.
- For each node, we recursively apply this check to its children, counting the good nodes: the node 3 has a good child node 1, and the node 1 has a good child node 1 (because 1<3 and 1<1 is not considered), and the node 4 has good child nodes 3 and 5 (because 3<4 and 5<4 is not considered, but 3<4 and 5>4).
- The total count of good nodes is 1+1+1+1=4, where each 1 represents a good node: the root node 3, and the nodes 1, 1, and 3.
Constraints:
- 1 <= number of nodes <= 10^5
Background Knowledge
The problem involves a binary tree, which is a fundamental data structure in computer science. A binary tree consists of nodes, where each node has at most two children (i.e., left child and right child). The topmost node is called the root. In this problem, we're given a binary tree represented as a level-order array, which means that the nodes are arranged in a specific order: the root node, followed by its children, then its grandchildren, and so on.
To solve this problem, we need to understand the concept of tree traversal, which refers to the process of visiting each node in the tree exactly once. There are several types of tree traversal, including in-order, pre-order, and post-order traversal. However, in this case, we're more interested in the path from the root to each node, as we need to check if there's a node with a value greater than the current node along this path.
The problem statement introduces the concept of a good node, which is a node that satisfies a specific condition: from the root to that node, there is no node with a value greater than it. This means that we need to keep track of the maximum value seen so far along the path from the root to each node. This is a classic example of a state-dependent problem, where the solution depends on the state of the system (in this case, the maximum value seen so far).
Algorithm/Approach
The general approach to solve this type of problem involves using a recursive or iterative approach to traverse the binary tree. We can use a depth-first search (DFS) or breadth-first search (BFS) algorithm to visit each node in the tree. Since we need to keep track of the maximum value seen so far along the path from the root to each node, we can use a state variable to store this information.
Step-by-Step Strategy
To solve this problem, we can follow these steps:
- Initialize a variable to store the count of good nodes.
- Define a recursive or iterative function to traverse the binary tree.
- For each node, check if it's a good node by comparing its value with the maximum value seen so far along the path from the root.
- If it's a good node, increment the count.
- Update the maximum value seen so far along the path from the root.
- Repeat the process for each node in the tree.
Common Pitfalls
When implementing the solution, we need to watch out for the following pitfalls:
- Forgetting to update the maximum value seen so far along the path from the root.
- Incorrectly checking if a node is a good node.
- Failing to handle the base case (e.g., an empty tree or a tree with a single node).
Time & Space Complexity
The expected time complexity for this problem is O(n), where n is the number of nodes in the tree, since we need to visit each node exactly once. The expected space complexity is O(h), where h is the height of the tree, since we need to store the recursive call stack or the state variable. In the worst case, the tree is skewed, and the space complexity becomes O(n).