Given a binary tree, determine whether it is a legal binary search tree (BST)
A BST is defined as:
The value in the left subtree of a node is strictly less than the value of the node.
The value in the node's right subtree is strictly greater than the value of the node.
The left and right subtrees must also be binary search trees.
A tree of nodes is also a binary search tree.
Two solutions:
recursion
class Solution(object):
def isValidBST(self, root, left=float("-inf"), right=float("inf")):
"""
当当前判断的节点在左边时，它应该比右上的父亲小，当当前判断的节点在右边时，它应该比左上的父亲大，否则返回False。
如果到叶子节点都未返回False的话，那么就全部合法，返回True即可
:type root: TreeNode
:rtype: bool
"""
if not root:
return True
if root.val >= right or root.val <= left:
return False
return self.isValidBST(root.left, left, min(right, root.val)) \
and self.isValidBST(root.right, max(left, root.val), right)
iterate
class Solution(object):
def isValidBST(self, root):
"""
迭代方式，定义跟节点及其节点应该对应的值范围，初始值范围设为[float('-inf'), float('inf')]
左侧节点取 min(父节点的值,上次范围大值)
右侧节点取 max(上次范围小值,父节点的值)
循环到当前节点只需要判断是否在此范围内即可
见下示例：中间中括号里的为树节点元素，左右为定义的合法范围
比如3下左节点1，它的范围右侧的值就是 min(inf, 3)
比如3下右节点5，它的范围左侧的值就是 max(-inf, 3)
-inf <[3]< inf
/ \
/ \
/ \
-inf <[1]< 3 3 <[5]< inf
/ \ / \
/ \ / \
/ \ / \
-inf<[0]<1 1<[2]<3 3<[4]<5 5<[6]<inf
:param root:
:return:
"""
if not root:
return True
p = [[root,float('-inf'), float('inf')]]
while len(p) > 0:
tmp = p.pop(0)
node = tmp[0]
less = tmp[1]
larger = tmp[2]
if less < node.val < larger:
pass
else:
return False
if node.left:
p.append([node.left, less, min(node.val, larger)])
if node.right:
p.append([node.right, max(less, node.val), larger])
return True