second-minimum-node-in-a-binary-tree.py

# Time:  O(n)
# Space: O(h)

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

class Solution(object):
    def findSecondMinimumValue(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def findSecondMinimumValueHelper(root, max_heap, lookup):
            if not root:
                return
            if root.val not in lookup:
                heapq.heappush(max_heap, -root.val)
                lookup.add(root.val)
                if len(max_heap) > 2:
                    lookup.remove(-heapq.heappop(max_heap))
            findSecondMinimumValueHelper(root.left, max_heap, lookup)
            findSecondMinimumValueHelper(root.right, max_heap, lookup)

        max_heap, lookup = [], set()
        findSecondMinimumValueHelper(root, max_heap, lookup)
        if len(max_heap) < 2:
            return -1
        return -max_heap[0]