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# Number of Good Leaf Nodes Pairs

Given the `root` of a binary tree and an integer `distance`. A pair of two different **leaf** nodes of a binary tree is said to be good if the length of **the shortest path** between them is less than or equal to `distance`.

Return *the number of good leaf node pairs* in the tree.

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/07/09/e1.jpg)

```
Input: root = [1,2,3,null,4], distance = 3
Output: 1
Explanation: The leaf nodes of the tree are 3 and 4 and the length of the shortest path between them is 3. This is the only good pair.
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/07/09/e2.jpg)

```
Input: root = [1,2,3,4,5,6,7], distance = 3
Output: 2
Explanation: The good pairs are [4,5] and [6,7] with shortest path = 2. The pair [4,6] is not good because the length of ther shortest path between them is 4.
```

**Example 3:**

```
Input: root = [7,1,4,6,null,5,3,null,null,null,null,null,2], distance = 3
Output: 1
Explanation: The only good pair is [2,5].
```

**Example 4:**

```
Input: root = [100], distance = 1
Output: 0
```

**Example 5:**

```
Input: root = [1,1,1], distance = 2
Output: 1
```

**Constraints:**

* The number of nodes in the `tree` is in the range `[1, 2^10].`
* Each node's value is between `[1, 100]`.
* `1 <= distance <= 10`

```java
class Solution {
    private int res;

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

    private int[] helper(TreeNode node, int distance) {
        if (node == null)
            return new int[11];

        // arr[i]: the number of leaf nodes at a distance 'i' from current node
        int[] A = new int[11];
        // Leaf Node
        if (node.left == null && node.right == null) {
            // Setting A[1] = 1, for the parent node
            A[1] = 1;
            return A;
        }
        int[] left = helper(node.left, distance);
        int[] right = helper(node.right, distance);
        // find all nodes satisfying distance
        for (int i = 0; i <= 10; ++i) {
            for (int j = 0; j <= 10; ++j) {
                if (i + j <= distance)
                    res += (left[i] * right[j]);
            }
        }
        // increment all by 1, ignore the node distance larger than 10
        for (int i = 0; i <= 9; ++i) {
            A[i + 1] += left[i];
            A[i + 1] += right[i];
        }
        return A;
    }
}
```


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