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1530. Number of Good Leaf Nodes Pairs

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:

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:

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

• Java
• C++
• Python
• Go

• /**
   * 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);
      }
  }
  

• /**
   * 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);
      }
  };
  

• # 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
  
  

• /**
   * 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)
  }
  

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