[LeetCode]671. Second Minimum Node In a Binary Tree

题目

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.

难度

Easy

方法

首先将最小值minimum和第二小值second_minimum置为maxint，递归遍历二叉树，如果节点值比最小值小，则将最小值赋给第二小值，然后将最小值更新为节点值；如果节点值比最小值大，并且节点值小于第二小值，则将第二小值更新为节点值。节点遍历后如果第二小值==maxint，即没有第二小值，则返回-1

python代码

import sys

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution(object):
    def findSecondMinimumValue(self, root):
        self.min = sys.maxint
        self.second_min = sys.maxint
        def traverse(node):
            if node:
                if node.val < self.min:
                    self.second_min = self.min
                    self.min = node.val
                elif node.val < self.second_min and node.val != self.min:
                    self.second_min = node.val

                traverse(node.left)
                traverse(node.right)

        traverse(root)
        if self.second_min != sys.maxint:
            return self.second_min
        return -1

root = TreeNode(2)
root.left = TreeNode(2)
root.right = TreeNode(5)
root.right.left = TreeNode(5)
root.right.right = TreeNode(7)

assert Solution().findSecondMinimumValue(root) == 5