#### determinate a BST tree

```
bool isBST(BinaryTree * p, int & min, int & max) {
if (!p) return true;
bool left = true;
bool right = true;
left = isBST(p - > left, min, max);
int currentMin = p - > left == null ? p - > data : min;
if (!left || p.data <= max) return false;
right = isBST(p - > right, min, max);
int currentMax = p - > right == null ? p - > data : max;
if (!right || p.data >= min) return false;
min = currentMin;
max= currentMax;
return true;
}
```

#### largest BST sub-tree in a Binary tree

// Find the largest BST subtree in a binary tree. // If the subtree is a BST, return total number of nodes. // If the subtree is not a BST, -1 is returned. int findLargestBSTSubtree(BinaryTree *p, int &min, int &max, int &maxNodes, BinaryTree *& largestBST) { if (!p) return 0; bool isBST = true; int leftNodes = findLargestBSTSubtree(p->left, min, max, maxNodes, largestBST); int currMin = (leftNodes == 0) ? p->data : min; if (leftNodes == -1 || (leftNodes != 0 && p->data <= max)) isBST = false; int rightNodes = findLargestBSTSubtree(p->right, min, max, maxNodes, largestBST); int currMax = (rightNodes == 0) ? p->data : max; if (rightNodes == -1 || (rightNodes != 0 && p->data >= min)) isBST = false; if (isBST) { // update min & max to pass to up level min = currMin; max = currMax; int totalNodes = leftNodes + rightNodes + 1; if (totalNodes > maxNodes) { maxNodes = totalNodes; largestBST = p; } return totalNodes; } else { return -1; // This subtree is not a BST } } BinaryTree* findLargestBSTSubtree(BinaryTree *root) { BinaryTree *largestBST = NULL; int min, max; int maxNodes = INT_MIN; findLargestBSTSubtree(root, min, max, maxNodes, largestBST); return largestBST; }

#### largest BST in a binary tree

Given a binary tree, find the largest Binary Search Tree (BST), where largest means BST with largest number of nodes in it. The largest BST may or may not include all of its descendants.

we use the term largest BST for largest BST which may or may not include all of its descendants, while largest BST subtree is for largest BST subtree which must include all of its descendants.

he largest BST subtree solution requires a bottom-up approach where the min and max values are passed bottom-up. The main reason of doing this is when one of the nodes does not satisfy the BST properties, all subtrees above (which includes this node as well) must also not satisfy the BST requirements.

However, finding the largest BST requires a slightly different approach. We want the largest BST by including as many nodes as possible while we traverse down the tree, as long the current BST constraint is maintained. we could not simply return root node of the largest BST as this would include all of its subtrees. You would need to create copies of the subtrees or delete nodes from the original binary tree.

```
A Top-down Approach:
// Find the largest BST in a binary tree.
// This code does not delete dynamically-allocated nodes,
// so memory will be leaked upon exit.
// The min and max values are passed top-down to check if
// including a node satisfies the current BST constraint.
// The child nodes are passed bottom-up to be assigned
// to its parent.
// Returns the total number of nodes the child holds.
int findLargestBST(BinaryTree *p, int min, int max, int &maxNodes,
BinaryTree *& largestBST, BinaryTree *& child) {
if (!p) return 0;
if (min < p->data && p->data < max) {
int leftNodes = findLargestBST(p->left, min, p->data, maxNodes, largestBST, child);
BinaryTree *leftChild = (leftNodes == 0) ? NULL : child;
int rightNodes = findLargestBST(p->right, p->data, max, maxNodes, largestBST, child);
BinaryTree *rightChild = (rightNodes == 0) ? NULL : child;
// create a copy of the current node and
// assign its left and right child.
BinaryTree *parent = new BinaryTree(p->data);
parent->left = leftChild;
parent->right = rightChild;
// pass the parent as the child to the above tree.
child = parent;
int totalNodes = leftNodes + rightNodes + 1;
if (totalNodes > maxNodes) {
maxNodes = totalNodes;
largestBST = parent;
}
return totalNodes;
} else {
// include this node breaks the BST constraint,
// so treat this node as an entirely new tree and
// check if a larger BST exist in this tree
findLargestBST(p, INT_MIN, INT_MAX, maxNodes, largestBST, child);
// must return 0 to exclude this node
return 0;
}
}
BinaryTree* findLargestBST(BinaryTree *root) {
BinaryTree *largestBST = NULL;
BinaryTree *child;
int maxNodes = INT_MIN;
findLargestBST(root, INT_MIN, INT_MAX, maxNodes, largestBST, child);
return largestBST;
}
```

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