#671

Second Minimum Node In a Binary Tree

pupil · 530 · lc easy +28 · verified · 46% accepted · 2,001 likes · top 30%

Description

You are given a non-empty special binary tree where every node has either exactly two or zero children. For nodes with two children, the parent's value equals the smaller of the two children's values — formally, root.val = min(root.left.val, root.right.val).

Find the second smallest distinct value across all nodes in the tree. If no such value exists, return -1.

Example 1:

Input: root = [2,2,5,null,null,5,7]

Output: 5

Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input: root = [2,2,2]

Output: -1

Explanation: The smallest value is 2, but there isn't any second smallest value.

Code

# 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 findSecondMinimumValue(self, root: Optional[TreeNode]) -> int:

# 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 findSecondMinimumValue(self, root: Optional[TreeNode]) -> int: