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

@TOC

# 题目描述

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.
``````

# 解题方法

# 找出所有值再求次小值

``````# 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.res = set()
self.inOrder(root)
if len(self.res) <= 1:
return -1
min1 = min(self.res)
self.res.remove(min1)
return min(self.res)

def inOrder(self, root):
if not root:
return
self.inOrder(root.left)
self.inOrder(root.right)
``````

# 遍历时求次小值

``````# 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
"""
if not root: return -1
self.res = float("inf")
self.min = root.val
self.inOrder(root)
return self.res if self.res != float("inf") else -1

def inOrder(self, root):
if not root:
return
self.inOrder(root.left)
if self.min < root.val < self.res:
self.res = root.val
self.inOrder(root.right)
``````

# 日期

2018 年 1 月 31 日 2018 年 11 月 19 日 —— 周一又开始了