Check if a tree is a Binary Search Tree (BST)

问题

I am trying to solve a binary search tree problem, but I can't pass all of the test cases. I need to return true if the tree is a binary search tree, otherwise, I need to return false. Can anyone tell me what I am doing wrong?

'''
class node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
'''

def checkBST(root):
    if root.left == None and root.right == None:
        return True
    if root == None:
        return True
    if root.left != None:
        if root.left.data < root.data:
            return True
    else:
        return False
    if root.right != None:
        if root.right.data > root.data:
            return True
        else:
            return False
    return chckBST(root.left) and chckBST(root) and chckBST(root.right)

回答1:

You've a lot of redundant if conditions in your code. You can simplify it like so:

def checkBST(root):
    if root == None or (root.left == None and root.right == None):
        return True

    elif root.right == None:
        return root.left.data < root.data and checkBST(root.left)

    elif root.left == None:
        return root.right.data >= root.data and checkBST(root.right)

    return checkBST(root.left) and checkBST(root.right)

First, check for all the None conditions. Short circuiting in python guarantees that if the first condition is False, the second will not be evaluated. This allows you to write succinct statements such as return root.left.data < root.data and checkBST(root.left).

Finally, in the event that neither left nor right nodes are None, do not call checkBST(root) again. This leads to infinite recursion.

回答2:

So the reason you're not passing some of the tests is because you're only checking one level deep. For instance, if there is a tree such that it has a root.left.right.data > root.data, then your code won't catch that. There is a good explanation here

But the gist is:

Your code will pass this

But it won't pass this Notice the right child of 2 > root.data

I think this solution solves it (sorry for answering a Python question with JS code, but I'm sure you'll get the idea):

function checkBST(root) {
    let isBST = true;
    let BSTUtil = r => {
        let left, right
        if(r.left) // Bottom out on the left side
            left = BSTUtil(r.left)
        if(r.right) // Bottom out on the right side
            right = BSTUtil(r.right)

        if(left > r.data) // Compare with parent
            isBST = false
        if(right < r.data) // Compare with parent
            isBST = false

        // Return MAX from branch
        if(!left && !right)
            return r.data
        else if(!left)
            return Math.max(right, r.data)
        else
            return Math.max(left, right, r.data)
    }
    BSTUtil(root)
    return isBST;
}

Also, please don't use this code, it uses O(n) space to solve the problem, and I'm sure I can find a more efficient solution if I spend some time on the question.

来源：https://stackoverflow.com/questions/44885472/check-if-a-tree-is-a-binary-search-tree-bst

标签

python

python-3.x

tree

binary-search-tree