Sum of Two Numbers in BSTs - Amazon Top Interview Questions


Problem Statement :


You are given two binary search trees a and b and an integer target. Return whether there's a number in a and a number in b such that their sum equals to target

Constraints

n ≤ 100,000 where n is the number of nodes in a
m ≤ 100,000 where m is the number of nodes in b

Example 1

Input

a = [5, [3, null, null], [7, null, null]]
b = [4, [2, null, null], [8, null, null]]

target = 9

Output

True

Explanation

We can pick 7 from a and 2 from b.

Example 2

Input

a = [5, [3, null, null], [7, null, null]]
b = [4, [2, null, null], [8, null, null]]

target = 4

Output

False


Solution :



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                        Solution in C++ :

bool solve(Tree* a, Tree* b, int target) {
    vector<Tree*> aStk;
    vector<Tree*> bStk;

    while ((a || !aStk.empty()) && (b || !bStk.empty())) {
        while (a) {
            aStk.push_back(a);
            a = a->left;
        }

        while (b) {
            bStk.push_back(b);
            b = b->right;
        }

        int sm = aStk.back()->val + bStk.back()->val;
        if (sm == target) return true;

        if (sm < target) {
            // need to increase sm
            a = aStk.back();
            aStk.pop_back();
            a = a->right;
        } else {
            b = bStk.back();
            bStk.pop_back();
            b = b->left;
        }
    }

    return false;
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, a, b, target):

        d = set()

        # iterate through a and save values
        s = []
        root = a
        while True:
            while root:
                s.append(root)
                root = root.left

            if not s:
                break

            # add target - val to our set
            node = s.pop()
            d.add(target - node.val)

            root = node.right

        # iterate through b and see if a value exists in our set
        s = []
        root = b
        while True:
            while root:
                s.append(root)
                root = root.left

            if not s:
                break

            node = s.pop()
            if node.val in d:
                return True

            root = node.right

        return False
                    

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