Skip to content

Latest commit

 

History

History
81 lines (67 loc) · 2.21 KB

File metadata and controls

81 lines (67 loc) · 2.21 KB

671. Second Minimum Node In a Binary Tree

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or zero sub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes. More formally, the property root.val = min(root.left.val, root.right.val) always holds.

Given such a binary tree, you need to output the second minimum value in the set made of all the nodes' value in the whole tree.

If no such second minimum value exists, output -1 instead.

Example 1:

Input:
    2
   / \
  2   5
     / \
    5   7
Output: 5
Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input:
    2
   / \
  2   2
Output: -1
Explanation: The smallest value is 2, but there isn't any second smallest value.

Solutions (Python)

1. Brute Force

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def findSecondMinimumValue(self, root: TreeNode) -> int:
        vals = set()
        nodes = [root]

        while nodes:
            curr = nodes.pop()
            vals.add(curr.val)
            if curr.left:
                nodes.append(curr.left)
                nodes.append(curr.right)

        vals.remove(min(vals))

        return min(vals) if vals else -1

2. Store 1st & 2nd Minimum Values

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def findSecondMinimumValue(self, root: TreeNode) -> int:
        ret = float('+inf')
        nodes = [root]

        while nodes:
            curr = nodes.pop()
            if root.val < curr.val < ret:
                ret = curr.val
            elif curr.val == root.val and curr.left:
                nodes.append(curr.left)
                nodes.append(curr.right)

        return ret if ret < float('+inf') else -1