Description#
You are 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://spcdn.pages.dev/leetcode/problems/1530.Number%20of%20Good%20Leaf%20Nodes%20Pairs/images/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://spcdn.pages.dev/leetcode/problems/1530.Number%20of%20Good%20Leaf%20Nodes%20Pairs/images/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].
Constraints:
- The number of nodes in the
tree
is in the range [1, 210].
1 <= Node.val <= 100
1 <= distance <= 10
Solutions#
Solution 1#
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| # Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def countPairs(self, root: TreeNode, distance: int) -> int:
def dfs(root, cnt, i):
if root is None or i >= distance:
return
if root.left is None and root.right is None:
cnt[i] += 1
return
dfs(root.left, cnt, i + 1)
dfs(root.right, cnt, i + 1)
if root is None:
return 0
ans = self.countPairs(root.left, distance) + self.countPairs(
root.right, distance
)
cnt1 = Counter()
cnt2 = Counter()
dfs(root.left, cnt1, 1)
dfs(root.right, cnt2, 1)
for k1, v1 in cnt1.items():
for k2, v2 in cnt2.items():
if k1 + k2 <= distance:
ans += v1 * v2
return ans
|
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| /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int countPairs(TreeNode root, int distance) {
if (root == null) {
return 0;
}
int ans = countPairs(root.left, distance) + countPairs(root.right, distance);
int[] cnt1 = new int[distance];
int[] cnt2 = new int[distance];
dfs(root.left, cnt1, 1);
dfs(root.right, cnt2, 1);
for (int i = 0; i < distance; ++i) {
for (int j = 0; j < distance; ++j) {
if (i + j <= distance) {
ans += cnt1[i] * cnt2[j];
}
}
}
return ans;
}
void dfs(TreeNode root, int[] cnt, int i) {
if (root == null || i >= cnt.length) {
return;
}
if (root.left == null && root.right == null) {
++cnt[i];
return;
}
dfs(root.left, cnt, i + 1);
dfs(root.right, cnt, i + 1);
}
}
|
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| /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int countPairs(TreeNode* root, int distance) {
if (!root) return 0;
int ans = countPairs(root->left, distance) + countPairs(root->right, distance);
vector<int> cnt1(distance);
vector<int> cnt2(distance);
dfs(root->left, cnt1, 1);
dfs(root->right, cnt2, 1);
for (int i = 0; i < distance; ++i) {
for (int j = 0; j < distance; ++j) {
if (i + j <= distance) {
ans += cnt1[i] * cnt2[j];
}
}
}
return ans;
}
void dfs(TreeNode* root, vector<int>& cnt, int i) {
if (!root || i >= cnt.size()) return;
if (!root->left && !root->right) {
++cnt[i];
return;
}
dfs(root->left, cnt, i + 1);
dfs(root->right, cnt, i + 1);
}
};
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| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func countPairs(root *TreeNode, distance int) int {
if root == nil {
return 0
}
ans := countPairs(root.Left, distance) + countPairs(root.Right, distance)
cnt1 := make([]int, distance)
cnt2 := make([]int, distance)
dfs(root.Left, cnt1, 1)
dfs(root.Right, cnt2, 1)
for i, v1 := range cnt1 {
for j, v2 := range cnt2 {
if i+j <= distance {
ans += v1 * v2
}
}
}
return ans
}
func dfs(root *TreeNode, cnt []int, i int) {
if root == nil || i >= len(cnt) {
return
}
if root.Left == nil && root.Right == nil {
cnt[i]++
return
}
dfs(root.Left, cnt, i+1)
dfs(root.Right, cnt, i+1)
}
|