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.
思路
在看到这道题的时候，我想起了二叉搜索树的一个特性就是二叉搜索树的中序遍历是升序排序的。因此根据这个特性。我得出二叉搜索树的中序遍历序列，然后从头到尾进行判断他是否是有序的。如果其中哪一个数字大小关系不满足，则说明不是有效的二叉搜索树。时间复杂度为O(n)，空间复杂度为O(n)。
解决代码
1 # Definition for a binary tree node.
2 # class TreeNode(object):
3 # def __init__(self, x):
4 # self.val = x
5 # self.left = None
6 # self.right = None
7
8 class Solution(object):
9 def isValidBST(self, root):
10 """
11 :type root: TreeNode
12 :rtype: bool
13 """
14 if not root:
15 return True
16 res = []
17 self.validBFS(root, res)
18 for i in range(1, len(res)): # 从头到尾进行判断是否存在不满足排序的元素。
19 if res[i] <= res[i-1]:
20 return False
21 return True
22
23
24 def validBFS(self, root, res): # 中序遍历函数，将遍历序列添加进去
25 if not root:
26 return
27 self.validBFS(root.left, res)
28 res.append(root.val)
29 self.validBFS(root.right, res)
非递归解决办法
1 # Definition for a binary tree node.
2 # class TreeNode(object):
3 # def __init__(self, x):
4 # self.val = x
5 # self.left = None
6 # self.right = None
7
8 class Solution(object):
9 def isValidBST(self, root):
10 """
11 :type root: TreeNode
12 :rtype: bool
13 """
14 if not root:
15 return True
16 res,stack = [], []
17
18 while stack or root:
19 if root:
20 stack.append(root)
21 root = root.left
22 else:
23 root = stack.pop()
24 if res and root.val <= res： # 每次添加前先判断
return False
26 res.append(root.val)
27 root = root.right
28 return True