[DFS/BFS]99클럽 코테 스터디 31일차 TIL + 1653-number-of-good-leaf-nodes-pairs

1653. Number of Good Leaf Nodes Pairs

Medium

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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

 

 

<내코드 호소>

# 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 inorder(root, leaf_list):
#             if root:
#                 inorder(root.left, leaf_list)
#                 if root.left is None and root.right is None:
#                     leaf_list.append(root)
#                 inorder(root.right, leaf_list)
#             return leaf_list
#         def dfs(root, leaf_list):
#             # visited = set()
#             distances = -1
#             stack = leaf_list
#             while stack:
#                 leaf = stack.pop()
#         leaf_list = []
#         leaf_list = inorder(root, leaf_list)
#         dfs(root, leaf_list)

class Solution:
    def countPairs(self, root: TreeNode, distance: int) -> int:
        
        def dfs(node):
            if not node:
                return []
            
            if not node.left and not node.right  : # falsy 처럼 조회
                return [1]

            left_distances = dfs(node.left)
            right_distances = dfs(node.right)

            for l in left_distances:
                for r in right_distances:
                    if l + r <= distance:
                        self.result += 1
            
            current_distances = [d + 1 for d in left_distances + right_distances if d + 1 <= distance]
            print(current_distances)

            return current_distances

        self.result = 0
        dfs(root)

        return self.result