[leetcode] 671. Second Minimum Node In a Binary Tree @ python

版权声明：版权归个人所有，未经博主允许，禁止转载 https://blog.csdn.net/danspace1/article/details/86564157

原题

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.

解法1

遍历二叉树, 将二叉树的值存储在列表里, 然后将列表除重并升序排列, 返回第二个元素, 如果第二个元素不存在则返回-1
Time: O(n), n为节点数
Space: O(1)

代码

# 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
        """
        l = []
        def getList(root):
            if not root:
                return
            getList(root.left)
            l.append(root.val)
            getList(root.right)
        getList(root)
        l = sorted(set(l))
        if len(l) >= 2:
            return l[1]
        return -1

解法2

参考: Python Extremely Easy To Understand (Beats 91%)
DFS. 观察二叉树可知, root是最小的节点, 因此我们深度优先搜索, 找到一个节点的值使得它比root的值大比其他所有节点值小, 每次遍历, 遇到符合条件的节点就更新self.ans, 如果遍历结束后self.ans不变则返回-1
Time: O(n)
Space: O(1)

代码

# 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
        """
        self.ans = sys.maxsize
        def dfs(node):
            if not node:
                return
            if root.val < node.val < self.ans:
                self.ans = node.val
            dfs(node.left)
            dfs(node.right)
            
        dfs(root)
        if self.ans == sys.maxsize:
            return -1
        return self.ans