Amount of Time for Binary Tree to Be Infected

2385. Amount of Time for Binary Tree to Be Infected

Medium

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You are given the root of a binary tree with unique values, and an integer start. At minute 0, an infection starts from the node with value start.

Each minute, a node becomes infected if:

  • The node is currently uninfected.

  • The node is adjacent to an infected node.

Return the number of minutes needed for the entire tree to be infected.

Example 1:

Input: root = [1,5,3,null,4,10,6,9,2], start = 3
Output: 4
Explanation: The following nodes are infected during:
- Minute 0: Node 3
- Minute 1: Nodes 1, 10 and 6
- Minute 2: Node 5
- Minute 3: Node 4
- Minute 4: Nodes 9 and 2
It takes 4 minutes for the whole tree to be infected so we return 4.

Example 2:

Input: root = [1], start = 1
Output: 0
Explanation: At minute 0, the only node in the tree is infected so we return 0.

Constraints:

  • The number of nodes in the tree is in the range [1, 10<sup>5</sup>].

  • 1 <= Node.val <= 10<sup>5</sup>

  • Each node has a unique value.

  • A node with a value of start exists in the tree.


Solution

# 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 amountOfTime(self, root: Optional[TreeNode], start: int) -> int:
        result = 0

        def DFS(node, start):
            if node == None:
                return 0

            leftDepth = DFS(node.left, start)
            rightDepth = DFS(node.right, start)

            if node.val == start:
                Solution.result = max(leftDepth, rightDepth)
                return -1

            elif leftDepth >= 0 and rightDepth >= 0:
                return max(leftDepth, rightDepth)+1

            Solution.result = max(Solution.result, abs(leftDepth - rightDepth))
            return min(leftDepth, rightDepth) - 1

        DFS(root, start)
        return Solution.result

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