/**
* Inorder successor of 8 is 10, inorder successor of 10 is 12 and inorder successor of 14 is 20.
*
* <p>Method 1 : Use parent pointer: If X has the right subtree, then successor must be the
* leftmodt child of the right subtree Else got to parent P. If X is on the left side then the
* next node is P If X is on right then we visited the P, we need to chk successor (P)
*/
public static BinaryTreeParent inordersucc1(BinaryTreeParent node) {
if (node == null) return null;
BinaryTreeParent parent;
// Found the right subtree
if (node.parent == null || node.right != null) parent = leftMostNode(node);
else {
// Go up to left side instead of right
while ((parent = node.parent) != null) {
if (parent.getLeft() == node) break;
node = parent;
}
}
return node;
}
public static BinaryTreeParent leftMostNode(BinaryTreeParent node) {
if (node == null) return null;
while (node != null) {
node = node.getLeft();
}
return node;
}
/**
* 1) If right subtree of node is not NULL, then succ lies in right subtree. Do following. Go to
* right subtree and return the node with minimum key value in right subtree. 2) If right sbtree
* of node is NULL, then start from root and us search like technique. Do following. Travel down
* the tree, if a nodeÕs data is greater than rootÕs data then go right side, otherwise go to left
* side.
*
* <p>Time complexity : O(h) // h is the height of the tree
*/
public static BinaryTreeParent inordersucc2(BinaryTreeParent node, BinaryTreeParent root) {
BinaryTreeParent succ = null;
if (node.getRight() != null) return leftMostNode(node);
while (root != null) {
if (node.data == root.data) return succ;
else if (node.data < root.data) {
succ = root;
root = root.getLeft();
} else {
root = root.getRight();
}
}
return succ;
}
