Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a 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:

    2
   / \
  1   3

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

Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [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.

解析

此題給予一二元樹，要求驗證該樹是否為二元搜尋樹。難度為Medium。

可透過recursion及iteration來解這道題。

解法一(Recursion)

時間複雜度：O(N)
空間複雜度：O(N)

class Solution:
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        def helper(node, lower = float('-inf'), upper = float('inf')):
            if not node:
                return True
            
            val = node.val
            if val <= lower or val >= upper:
                return False

            if not helper(node.right, val, upper):
                return False
            if not helper(node.left, lower, val):
                return False
            return True

        return helper(root)

解法二(Iteration)

時間複雜度：O(N)
空間複雜度：O(N)

class Solution:
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
            
        stack = [(root, float('-inf'), float('inf')), ] 
        while stack:
            root, lower, upper = stack.pop()
            if not root:
                continue
            val = root.val
            if val <= lower or val >= upper:
                return False
            stack.append((root.right, val, upper))
            stack.append((root.left, lower, val))
        return True  

備註

希望透過記錄解題的過程，可以對於資料結構及演算法等有更深一層的想法。
如有需訂正的地方歡迎告知，若有更好的解法也歡迎留言，謝謝。