[LeetCode] DFS 98. Validate Binary Search Tree

문제

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

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코드

// DFS

var isValidBST = function(root, min = -Infinity, max = Infinity) {
	// root가 없는 경우는
    // 그래프가 아예 없거나, left, right 둘 다 없는 경우임
    // 둘다 없을 경우 해당 root.val는 true
    if(!root) return true;
    
    // root.val은 max보다는 작아야하고
    // min보다는 커야함
    if(root.val >= max || root.val <= min) return false;

	// left는 val보다 작고 min보다는 큼
    const left = isValidBST(root.left, min, root.val);

	// right는 val보다 크고 max보다는 작음
	const right = isValidBST(root.right, root.val, max);
    
    return left && right
}