LeetCode 1080. Insufficient Nodes in Root to Leaf Paths Solution in Java, C++, Python & More | Explanation + Code

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1080. Insufficient Nodes in Root to Leaf Paths

Description

Given the root of a binary tree and an integer limit, delete all insufficient nodes in the tree simultaneously, and return the root of the resulting binary tree.

A node is insufficient if every root to leaf path intersecting this node has a sum strictly less than limit.

A leaf is a node with no children.

 

Example 1:

Input: root = [1,2,3,4,-99,-99,7,8,9,-99,-99,12,13,-99,14], limit = 1
Output: [1,2,3,4,null,null,7,8,9,null,14]

Example 2:

Input: root = [5,4,8,11,null,17,4,7,1,null,null,5,3], limit = 22
Output: [5,4,8,11,null,17,4,7,null,null,null,5]

Example 3:

Input: root = [1,2,-3,-5,null,4,null], limit = -1
Output: [1,null,-3,4]

 

Constraints:

  • The number of nodes in the tree is in the range [1, 5000].
  • -105 <= Node.val <= 105
  • -109 <= limit <= 109

Solutions

Solution 1

PythonJavaC++GoTypeScriptJavaScript
# 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 sufficientSubset( self, root: Optional[TreeNode], limit: int ) -> Optional[TreeNode]: if root is None: return None limit -= root.val if root.left is None and root.right is None: return None if limit > 0 else root root.left = self.sufficientSubset(root.left, limit) root.right = self.sufficientSubset(root.right, limit) return None if root.left is None and root.right is None else root(code-box)

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