**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 :

` ````
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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