671. Second Minimum Node In a Binary Tree 二叉树中第二小节点

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly twoor 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.

给定由非负值的节点组成的非空特殊二叉树，其中该树中的每个节点都具有两个或零个子节点。如果节点有两个子节点，则该节点的值是其两个子节点之间的较小值。 给定这样一个二叉树，您需要输出整个树中所有节点值的集合中的第二个最小值。 如果不存在这样的第二个最小值，则输出-1。

# 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
        """
        s = set()
        queue = [root]
        while queue and queue[0]:
            node = queue.pop(0)
            s.add(node.val)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
        if len(s) < 2:
            return -1
        else:
            s.remove(min(s))
            return min(s)

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