98. Validate Binary Search Tree[Medium]

98. Validate Binary Search Tree

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.

Constraints:

• The number of nodes in the tree is in the range [1, 10^4].
• -2^31 <= Node.val <= 2^31 - 1

Example

Input: root = [2,1,3]
Output: true

思路

BST就是要保证左子树的值小于当前节点值，右子树的值大于当前节点值。
换句话说就是每一个节点的值都要小于某个值并大于某个值，那根节点就大于MIN_VALUE，小于MAX_VALUE，递归每一个节点

• 如果是左子树，那就应该小于父节点的value，大于父节点的left
• 如果是右子树，那就应该大于父节点的value，小于父节点的right

题解

    public boolean isValidBST(TreeNode root) {
	// 要用Long.类型最大值最小值，题目限制最大值是2^31-1，Integer最大值不够
        return dfsIsValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }

    public boolean dfsIsValidBST(TreeNode root, long left, long right) {
        if (root == null)
            return true;
        if (!(root.val > left && root.val < right))
            return false;

        return dfsIsValidBST(root.left, left, root.val) && dfsIsValidBST(root.right, root.val, right);
    }