1530. Number of Good Leaf Nodes Pairs

Intuition

To find the number of good leaf node pairs in a binary tree, we can use Depth-First Search (DFS) to explore the tree. By tracking the distance of leaf nodes from the current node, we can determine if two leaf nodes form a good pair. A key insight is that we only need to check pairs of leaf nodes from different subtrees because pairs from the same subtree are not good by definition.

Approach: DFS

Explanation:

1. Base Case for Leaf Nodes:
2. If a node is null, return an empty vector as there are no distances to track.
3. If a node is a leaf (no children), return a vector containing a single element 1, which represents the distance from this leaf node to itself.
4. DFS on Left and Right Subtrees:
5. Recursively perform DFS on the left and right subtrees of the current node.
6. Obtain vectors left and right which contain the distances of leaf nodes in the left and right subtrees from the current node.
7. Count Good Pairs:
8. For each pair of distances from the left and right vectors, check if their sum is less than or equal to the given distance. If it is, increment the result counter res.
9. Update Distances and Return:
10. Create a new vector distanceList to store updated distances.
11. Increment each distance from the left vector by 1 (to account for the current node) and add it to distanceList if it is less than the given distance.
12. Similarly, increment each distance from the right vector by 1 and add it to distanceList.
13. Return distanceList for use in the parent node's computations.

Complexity

• Time complexity: O(n²), where n is the number of nodes in the tree.
• Space complexity: O(n)

Code

class Solution {
private:
    vector<int> dfs(TreeNode* root, int& distance, int& res) {
        if (!root) return {};
        if (!root->left && !root->right) return {1};

        auto left = dfs(root->left, distance, res);
        auto right = dfs(root->right, distance, res);

        for (const auto& distanceLeft: left) {
            for (const auto& distanceRight: right) {
                if (distanceLeft + distanceRight <= distance) {
                    ++res;
                }
            }
        }

        vector<int> distanceList;
        for (auto& distanceLeft: left) {
            if (++distanceLeft < distance)
                distanceList.push_back(distanceLeft);
        }
        for (auto& distanceRight: right) {
            if (++distanceRight < distance)
                distanceList.push_back(distanceRight);
        }

        return distanceList;
    }

public:
    int countPairs(TreeNode* root, int distance) {
        int res = 0;
        dfs(root, distance, res);
        return res;
    }
};