LeetCode---Maximum Binary Tree、Second Minimum Node In a Binary Tree、Search in a Binary Search Tree

654. Maximum Binary Tree

Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:

1. The root is the maximum number in the array.
2. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
3. The right subtree is the maximum tree constructed from right part subarray divided by the maximum number.

 

Construct the maximum tree by the given array and output the root node of this tree.

Example 1:

Input: [3,2,1,6,0,5]
Output: return the tree root node representing the following tree:

      6
    /   \
   3     5
    \    / 
     2  0   
       \
        1

给定一个不含重复元素的整数数组。一个以此数组构建的最大二叉树定义如下：

二叉树的根是数组中的最大元素。
左子树是通过数组中最大值左边部分构造出的最大二叉树。
右子树是通过数组中最大值右边部分构造出的最大二叉树。
通过给定的数组构建最大二叉树，并且输出这个树的根节点。

思路：先找到数组最大值，设置为根节点，然后进行递归

# 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 constructMaximumBinaryTree(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        def getTree(nums,start,end):
            if start==end:
                return None
            ma=max(nums[start:end])
            node=TreeNode(ma)
            node.left=getTree(nums,start,nums.index(ma))
            node.right=getTree(nums,nums.index(ma)+1,end)
            return node
        return getTree(nums,0,len(nums))

671. Second Minimum Node In a Binary Tree

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. More formally, the property root.val = min(root.left.val, root.right.val) always holds.

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.

给定一个非空特殊的二叉树，每个节点都是正数，并且每个节点的子节点数量只能为 2 或 0。如果一个节点有两个子节点的话，那么这个节点的值不大于它的子节点的值。 

给出这样的一个二叉树，你需要输出所有节点中的第二小的值。如果第二小的值不存在的话，输出 -1 。

思路：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

# 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
        """
        res=[]
        def getlist(root):
            if root is None:
                return
            getlist(root.left)
            res.append(root.val)
            getlist(root.right)
        getlist(root)
        res=sorted(set(res))
        if len(res)>=2:
            return res[1]
        return -1

700. Search in a Binary Search Tree

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.

For example, 

Given the tree:
        4
       / \
      2   7
     / \
    1   3

And the value to search: 2

You should return this subtree:

      2     
     / \   
    1   3

In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

给定二叉搜索树（BST）的根节点和一个值。 你需要在BST中找到节点值等于给定值的节点。 返回以该节点为根的子树。 如果节点不存在，则返回 NULL。

思路：树的查找比较简单的就是递归，如果节点是空，那么肯定没找到；节点值相等，返回这个节点；如果节点值小于要查找的值，那么在当前节点的右子树中找；否则在左子树中找。

# 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 searchBST(self, root, val):
        """
        :type root: TreeNode
        :type val: int
        :rtype: TreeNode
        """
        if root is None:
            return None
        if root.val==val:
            return root
        elif root.val<val:
            return self.searchBST(root.right,val)
        else:
            return self.searchBST(root.left,val)