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Why the solution for binary search tree validity is not working?


Why the solution for binary search tree validity is not working?

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Why the solution for binary search tree validity is not working?
Tag : java , By : ffmmjj
Date : November 28 2020, 09:01 AM

I wish this help you I wrote a solution for checking the validity of the binary search tree using LinkedList and it's proving wrong information about the validity. I checked with a valid BST and it returns that the tree is not valid. The code is as following, , You seem to have your checks the wrong way around:
if ( ( cur.left != null && cur.data > cur.left.data ) || (cur.right != null && cur.data < cur.right.data  ) )
boolean isValid(Node node) {
    if (node.left != null && (cur.data < node.left.data || !isValid(node.left)))
        return false;
    else if (node.right != null && (cur.data > node.right.data || !isValid(node.right)))
        return false;
    else
        return true;
}
public boolean isValid(Node root) {
    LinkedList<Node> nodesToCheck = new LinkedList<>();
    nodesToCheck.offer(root);
    while (!nodesToCheck.isEmpty()) {
        Node current = nodesToCheck.poll();
        if (current.left != null) {
            if (current.data < current.left.data)
                return false;
            nodesToCheck.offer(current.left);
        }
        if (current.right != null) {
            if (current.data > current.right.data)
                return false;
            nodesToCheck.offer(current.right);
        }
    }
    return true;
}

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Where to Add Checks for Validity in a Binary Search Tree


Tag : java , By : Michael
Date : March 29 2020, 07:55 AM
it should still fix some issue Your class will have contracts. A reasonable expectation for a binary search tree would be the contract that every such tree is indeed a binary search tree. This is called an invariant.
Every operation that manipulates such a tree should do it in such a way that this invariant is never broken.

Is binary tree a binary search tree if tree is spread over multiple machines


Tag : algorithm , By : Bharath
Date : March 29 2020, 07:55 AM
around this issue BST has a property. it's each children will also be a BST. validate all the machine's binary tree and once you have the each machine BT is BST then get the root node of each machine's BT and then again validate the tree if it is BST from the root node.

JavaScript Class solution for Validate Binary Search Tree


Tag : javascript , By : Scott Everts
Date : March 29 2020, 07:55 AM
will be helpful for those in need Thank you guys for your answers. Now I understand the issue: LeetCode forces me to run the function isValidBST that takes the parameter root and return the result. Doing so solves it:
class Solution {
    constructor(root) {
        this.root = root;
    }

    get result() {
        return this.helper(this.root, -Infinity, Infinity);
    }

    helper(root, low, high) {
        if (!root) return true;
        else {
            let val = root.val;
            if (val <= low || val >= high) return false;
            if (!this.helper(root.right, val, high)) return false;
            if (!this.helper(root.left, low, val)) return false;
            return true;
        }
    }
}

var isValidBST = function(root) {
    const res = new Solution(root)
    return res.result;
};

Differences in implementation of a tree, binary tree and, binary search tree in python


Tag : python , By : suresh
Date : March 29 2020, 07:55 AM

Function to check whether a binary tree is binary search tree or not working


Tag : cpp , By : dbarbot
Date : March 29 2020, 07:55 AM
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