**Difficulty:** Easy, **Asked-in:** Facebook, Google, Amazon, Microsoft.

**Key takeaway:** An excellent problem to learn problem-solving using both iterative and recursive approaches in the binary search tree.

Given a node x and the root of a binary search tree, write a program to find the in-order successor of node x.

- The in-order successor of node x is a node that will be visited just after node x during in-order traversal. In other words, the in-order successor of a node x is the node with the smallest key greater than x->key.
- If node x is visited last during the in-order traversal, then its in-order successor will be NULL. In other words, such a node will be the node with the largest key in the BST.

To find the in-order successor, we need to find the node with the smallest key just greater than the x->key. So there will two cases:

**Case 1: When the right subtree of x is not empty**

In this case, the in-order successor of node x is the node with the minimum key in its right subtree, i.e., the leftmost node in the right subtree of node x.

```
if (x->right != NULL)
return bstMinimum(x->right)
```

**Case 2: When the right subtree of x is empty**

In this case, the in-order successor is the lowest ancestor of x whose left child is also an ancestor of x. To find such an ancestor, we move up in the tree from node x until we encounter a node that is the left child of its parent. If any such node is found, then the parent of that node is the in-order successor. Otherwise, an in-order successor does not exist for the node (if x is the node with the largest key).

Here, we need to move upward towards the ancestors of the node. So, the implementation will be easier if we define a BST node structure that has a pointer to its parent as well.

```
BSTNode prev = x->parent
while (x != NULL && x == prev->right)
{
x = prev
prev = prev->parent
}
return prev
```

```
struct BSTNode
{
int key;
BSTNode* left;
BSTNode* right;
BSTNode* parent;
BSTNode(int k)
{
key = k;
left = NULL;
right = NULL;
parent = NULL;
}
};
BSTNode* bstMinimum(BSTNode* root)
{
BSTNode* curr = root;
while (curr->left != NULL)
curr = curr->left;
return curr;
}
BSTNode* bstInorderSuccessor(BSTNode* root, BSTNode* x)
{
if (x->right != NULL)
return bstMinimum(x->right);
BSTNode* prev = x->parent;
while (x != NULL && x == prev->right)
{
x = prev;
prev = prev->parent;
}
return prev;
}
```

```
class BSTNode:
def __init__(self, key):
self.key = key
self.left = None
self.right = None
self.parent = None
def bst_minimum(root):
curr = root
while curr.left is not None:
curr = curr.left
return curr
def bst_inorder_successor(root, x):
if x.right is not None:
return bst_minimum(x.right)
prev = x.parent
while x is not None and x == prev.right:
x = prev
prev = prev.parent
return prev
```

If the right child of node x is present, then the minimum value in the right subtree will be the successor. In the worst case, node x will be the root node, and we need to traverse the entire height of the tree down to one of the deepest leaf nodes. So, the time complexity in the worst case is O(h).

If the right child of x is not present, then we traverse upward to find the successor. In the worst case, x will be one of the deepest leaf nodes, and the root node will be the successor. Here, we need to traverse the entire height of the tree. So, the time complexity in the worst case is O(h).

Overall, the time complexity is O(h), which depends on the height of the tree. Since we store an extra parent pointer inside each node, the space complexity is O(n).

In this approach, the implementation of the first case remains the same. When the right subtree of x is not empty, the in-order successor of node x is the node with the minimum key in its right subtree.

For case 2, when the right subtree of x is empty, we move down from the root to node x and update the in-order successor. The idea is simple: We maintain a **succ** pointer to track the successor and move down from the root to node x by comparing the root->key with the x->key.

**If x->key > root->key**, we move right. There is no need to update the**succ**pointer because the current node root will not be the candidate for the successor. The idea is simple: the value of**succ->key**will always be greater than**x->key**.**If x->key < root->key**, we update the**succ**pointer to the current node**root**and move to the left subtree (**succ = root**and**root = root->left**). We update the**succ**pointer because the current node root will be the candidate for the successor. Think!**If x->key == root->key**, it means we have found the given node x, and we break out of the loop.

At the end of this loop, we return the **succ** pointer, which will hold the successor of the node x.

```
BSTNode* bstMinimum(BSTNode* root)
{
BSTNode* curr = root;
while (curr->left != NULL)
curr = curr->left;
return curr;
}
BSTNode* bstInorderSuccessor(BSTNode* root, BSTNode* x)
{
if (x->right != NULL)
return bstMinimum(x->right);
BSTNode* succ = NULL;
BSTNode* curr = root;
while (curr != NULL)
{
if (x->key < curr->key)
{
succ = curr;
curr = curr->left;
}
else if (x->key > curr->key)
curr = curr->right;
else
break;
}
return succ;
}
```

```
def bst_minimum(root):
curr = root
while curr.left is not None:
curr = curr.left
return curr
def bst_inorder_successor(root, x):
if x.right is not None:
return bst_minimum(x.right)
succ = None
curr = root
while curr is not None:
if x.key < curr.key:
succ = curr
curr = curr.left
elif x.key > curr.key:
curr = curr.right
else:
break
return succ
```

During the search process, we traverse the entire height of the tree in the worst case. So, the time complexity will be O(h). Since there is no parent pointer and we use a constant amount of extra space, the space complexity is O(1).

The idea is to search for the given node in the tree and update the successor to the current node before visiting its left subtree. If the node is found in the BST, we return the node with the minimum key in its right subtree. If the right subtree of the node doesn't exist, then the in-order successor is one of its ancestors, which has already been updated while searching for the given key.

**If root->key == x->key**, we check if node x has the right subtree. If it does, we return the minimum node in the right subtree as the successor, i.e., return**bstMinimum(root->right)**.- If root->key > x->key, we update the succ pointer to the current node root and recursively call the same function on the left subtree, i.e.,
**inorderSuccessor(root->left, succ, x)**. - If root->key < x->key, we recursively call the same function on the right subtree to continue searching for the successor in the right subtree, i.e.,
**inorderSuccessor(root->right, succ, x)**. - Finally, we return the succ pointer.

```
BSTNode* bstMinimum(BSTNode* root)
{
while (root->left != NULL)
root = root->left;
return root;
}
BSTNode* inorderSuccessor(BSTNode* root, BSTNode* succ, BSTNode* x)
{
if (root == NULL)
return succ;
if (x->key == root->key)
{
if (x->right != nullptr)
return bstMinimum(x->right);
}
else if (x->key < root->key)
{
succ = root;
return inorderSuccessor(root->left, succ, x);
}
else
return inorderSuccessor(root->right, succ, x);
return succ;
}
```

```
def bst_minimum(root):
while root.left is not None:
root = root.left
return root
def inorder_successor(root, succ, x):
if root is None:
return succ
if x.key == root.key:
if x.right is not None:
return bst_minimum(x.right)
elif x.key < root.key:
succ = root
return inorder_successor(root.left, succ, x)
else:
return inorder_successor(root.right, succ, x)
return succ
```

At each step of recursion, we descend one level down. So, in the worst case, we traverse the entire height of the tree. Time complexity = O(h).

The space complexity depends on the size of the call stack, which is equivalent to the height of the recursion tree. So, the space complexity is O(h).

- How can we solve this problem using an in-order traversal?
- How can we modify the above code to find both in-order predecessor and successor together?
- How can we solve this problem by storing the in-order traversal of the BST in an array?
- The time complexity of all the above approaches is O(h). What would be the worst and best-case scenarios?

- Replace each node in the binary tree with the sum of its inorder predecessor and successor
- Populate inorder successor for all nodes
- Find the in-order predecessor of a node in BST
- The in-order successor of a node in a binary tree
- The lowest common ancestor of a binary search tree
- Validate binary search tree
- Convert sorted array to binary search tree

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