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ValidateBinarySearchTree.scala
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ValidateBinarySearchTree.scala
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package ValidateBinarySearchTree
/**
* Created by iodone on {19-10-16}.
*/
class TreeNode(var _value: Int) {
var value: Int = _value
var left: TreeNode = null
var right: TreeNode = null
}
/**
* 解题报告
*
* 思路:BST的判断不是与构建的时候那么明显,特别需要注意每个节点是有上界和下界的。在处理left节点容易忽略上界,right节点容易忽略下界。
* 采用递归思路。通过min和max保存当前节点的上下界,同时进入到左子树时更新上界,进入到右子树更新下界。
*
*/
object Solution {
def isValidBST(root: TreeNode): Boolean = {
// inorderTraversal(root)._2
isValid(root, BigInt(Int.MinValue)-1, BigInt(Int.MaxValue)+1)
}
def inorderTraversal(root: TreeNode): (List[Int], Boolean) = {
if (root == null) (List(), true)
else {
val left = inorderTraversal(root.left)
val right = inorderTraversal(root.right)
val isValid = (left._2 && right._2) && (left._1.isEmpty || left._1.last < root.value) && (right._1.isEmpty || right._1.head > root.value)
(left._1 ::: root.value :: right._1, isValid)
}
}
def isValid(root: TreeNode, min: BigInt, max: BigInt): Boolean = {
if (root == null) true
else {
if (root.value > min && root.value < max) {
isValid(root.left, min, root.value) && isValid(root.right, root.value, max)
}
else false
}
}
def main(args: Array[String]): Unit = {
val node0 = new TreeNode(5)
val node1 = new TreeNode(4)
val node2 = new TreeNode(7)
node0.left = node1
node0.right = node2
println(isValidBST(node0))
}
}