98. Validate Binary Search Tree
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
- The left subtree of a node contains only nodes with keys less than
the node’s key. - The right subtree of a node contains only nodes with keys greater
than the node’s key. - Both the left and right subtrees must also be binary search trees.
Example 1:
2
/ \
1 3
Input: [2,1,3]
Output: true
Example 2:
5
/ \
1 4
/ \
3 6
Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.
tree的定义：val、left、right
二叉搜索树有效是指每个节点的值大于左节点，小于右节点，且它的左节点和右节点也满足这种条件。
注意对于每个节点的判断，是要给定一个最小值和最大值的，因为中间节点必须大于左侧所有子节点，小于右侧所有子节点（注意小于，不能等于）。
用float(‘inf’)定义最值，所以递归的时候要传入两个值。
# 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 isValidBST(self, root, lowest = float('-inf'), highest = float('inf')):
"""
:type root: TreeNode
:rtype: bool
"""
if root == None:
return True
val = root.val
if val <= lowest or val>= highest:
return False
if not self.isValidBST(root.left, lowest, val):
return False
if not self.isValidBST(root.right, val, highest):
return False
return True
第二种解法：使用中序遍历，基本上所有二叉搜索树的操作都可以这样做
四种主要的遍历思想为：
前序遍历：根结点 —> 左子树 —> 右子树
中序遍历：左子树—> 根结点 —> 右子树
后序遍历：左子树 —> 右子树 —> 根结点
层次遍历：只需按层次遍历即可
注意：pre节点要定义在class内，定义的时候没有self
# 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):
pre=None
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if root is None:
return True
Bool= self.isValidBST(root.left)
if self.pre!=None:
Bool=Bool and (self.pre.val<root.val)
self.pre=root
Bool=Bool and self.isValidBST(root.right)
return Bool
99. Recover Binary Search Tree
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Example 1:
Input: [1,3,null,null,2]
1
/
3
\
2
Output: [3,1,null,null,2]
3
/
1
\
2
Example 2:
Input: [3,1,4,null,null,2]
3
/ \
1 4
/
2
Output: [2,1,4,null,null,3]
2
/ \
1 4
/
3
Follow up:
A solution using O(n) space is pretty straight forward.
Could you devise a constant space solution?
同样使用中序遍历即可，所以还是需要定义pre节点表示前驱节点。
有两个节点颠倒了，只需要找到两个节点的位置。用n1存储第一个错位的元素，n2存储第二个，然后交换就可以了。
# 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 recoverTree(self, root):
"""
:type root: TreeNode
:rtype: None Do not return anything, modify root in-place instead.
"""
self.n1, self.n2, self.pre = None, None, None
self.inorder(root)
self.n1.val, self.n2.val = self.n2.val, self.n1.val
def inorder(self, root):
if root ==None:
return
self.inorder(root.left)
if self.pre!=None and self.pre.val > root.val:
if self.n1 == None:
self.n1=self.pre
self.n2=root
self.pre=root
self.inorder(root.right)