Validate Binary Search Tree(iterative,dfs)

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:

Input:

    2
   / \
  1   3

Output:
 true

Example 2:

    5
   / \
  1   4
     / \
    3   6

Output:
 false

Explanation:
 The input is: [5,1,4,null,null,3,6]. The root node's value
             is 5 but its right child's value is 4.

分析

中序遍历 while

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        stack = []
        cur = root
        prev = None
        while cur or stack:
            while cur:
                stack.append(cur)
                cur = cur.left
            cur = stack.pop()
            if prev and cur.val<=prev.val:
                return False
            prev = cur
            cur = cur.right
        return True

dfs

用root.val 传入dfs参数，作为最大值最小值

float('-inf'),float('inf')

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        return self.helper(root,float('-inf'),float('inf'))

    def helper(self,root,minn,maxx):
        if not root:
            return True

        if root.val >= maxx or root.val <= minn:
            return False
        return self.helper(root.left,minn,root.val) and self.helper(root.right,root.val,maxx)