Given the root of a binary tree, determine if it is a valid binary search tree (BST).
A valid 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:
Input: root = [2,1,3]
Output: true
Example 2:
Input: root = [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.
Constraints:
The number of nodes in the tree is in the range [1, 104].
-231 <= Node.val <= 231 - 1
Solutions
Solution 1: Recursion
We can perform a recursive in-order traversal on the binary tree. If the result of the traversal is strictly ascending, then this tree is a binary search tree.
Therefore, we use a variable prev to save the last node we traversed. Initially, prev = -∞. Then we recursively traverse the left subtree. If the left subtree is not a binary search tree, we directly return False. Otherwise, we check whether the value of the current node is greater than prev. If not, we return False. Otherwise, we update prev to the value of the current node, and then recursively traverse the right subtree.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the binary tree.
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# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclassSolution:defisValidBST(self,root:Optional[TreeNode])->bool:defdfs(root:Optional[TreeNode])->bool:ifrootisNone:returnTrueifnotdfs(root.left):returnFalsenonlocalprevifprev>=root.val:returnFalseprev=root.valreturndfs(root.right)prev=-infreturndfs(root)
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcisValidBST(root*TreeNode)bool{varprev*TreeNodevardfsfunc(*TreeNode)booldfs=func(root*TreeNode)bool{ifroot==nil{returntrue}if!dfs(root.Left){returnfalse}ifprev!=nil&&prev.Val>=root.Val{returnfalse}prev=rootreturndfs(root.Right)}returndfs(root)}