Algo#41: Free given binary tree.

Given problem of freeing data structure checks whether you know in & out of given data structure or not. You can verify memory leaks using Valgrind.

Following is implementation of above problem in c language.

Please write comments if you find anything wrong or you want to add something more related to this topic.

Algo#31: Find nth node in inorder traversal.

Given problem of finding nth node in in-order traversal can be easily done if we keep count of nde visited while doing in-order traversal. We will do keep this counter as static rather than global. Though taking static variable are also not encouraged in general because of parallel programming. But as of now in our case it will do.

Following is implementation of above algorithm in c language.

Please write comments if you find anything wrong or you want to add something more related to this topic.

Algo#30: Iterative implementation of Post-order, Pre-order & In-order traversal of given binary tree

We mostly use recursive function whenever we are asked traversal of binary tree. But what if you have to do it iterative. All recursion function can be converted to iterative with more or less effort.

Here we will see Post-order, Pre-order & In-order traversal of given binary tree using iterative method.

Following is implementation of above problem in c language.

Please write comments if you find anything wrong or you want to add something more related to this topic.

Algo#7: Closest Value Search in Binary Search Tree (BST)

Closest Value Search in Binary Search Tree (BST) means returning nearest value in BST to given key.

Following is simple algorithm for finding Closest Value in BST.

Step 1: Current root itself is NULL, then closest value is zero.
Step 2: If current root matches target key value, then closest value is key itself.
Step 3: Otherwise, consider current root as closest value and do following.
        3.1: If key is smaller then current root, find closest value on left side tree of current root recursively and call it RecursiveClosest.
        3.2: If key is larger then current root, find closest value on right side tree of current root recursively and call it RecursiveClosest.
Step 4: Return current root or RecusiveClosest depending on which ever is nearer to target key.

Following is implementation of above algorithm in c language.

Please write comments if you find anything wrong or you want to add something more related to this topic.

Algo#6: Preorder Successor in Binary Tree

Preorder Successor value for any node X means value of node Y that comes just after node X while doing Preorder traversal. For given tree in figure, preorder traversal (Root Left Right) would be : 1 2 4 5 3 6. We can search preorder successor for any node by just looking at this traversal output. E.G. preorder successor of node '3' is node '6' because '6' appears after '3' in given traversal.

But in real life we encounter Preorder Successor to be search for any node randomly, and if we go by this method of doing preorder traversal first then giving successor, it would be inefficient because of its time complexity of O(n) in large trees. So we need some method which can return Preorder Successor randomly for any node without doing traversal explicitly in less time complexity.

Luckily preorder traversal follows some pattern which can help us in giving Preorder Successor for any node without doing traversal at all.

Following is simple algorithm for finding Preorder Successor of binary tree.

Step 1: Current root itself is NULL, then successor is also NULL.

Step 2: Current root contains value same as key for which we are looking successor.

        2.1: Current root has right child, then left most node of right child is successor.

        2.2: Current root does have right child r, then r is successor.
        2.3: Current root is left leaf node and having sibling on right side

        2.4: Otherwise, if current root has an ancestor, v, which is a left-child and v has a right sibling, vrs, then succ(current root) is vrs
        2.5: If none of above applies, then succ(current root) doesn't exist.

Step 3: Current root is not the target node for which we are looking successor.

        3.1: Search target node and its successor in left side of tree recursively, and return if found.

        3.2: Search target node and its successor in right side of tree recursively, and return.

Following is implementation of above algorithm in c language.

Please write comments if you find anything wrong or you want to add something more related to this topic.

Algo#5: Preorder Predecessor in Binary Tree

Preorder Predecessor value for any node X means value of node Y that comes just before node X while doing Preorder traversal. For given tree in figure, preorder traversal (Root Left Right) would be : 1 2 4 5 3 6. We can search preorder predecessor for any node by just looking at this traversal output. E.G. preorder predecessor of node '4' is node '2' because '2' appears before '4' in given traversal.

But in real life we encounter Preorder Predecessor to be search for randomly any node, and if we go by this method of doing preorder traversal first then giving predecessor, it would be inefficient because of its time complexity of O(n) in large trees. So we need some method which can return Preorder Predecessor randomly for any node without doing traversal explicitly in less time complexity.

Luckily preorder traversal follows some pattern which can help us in giving Preorder Predecessor for any node without doing traversal at all.

Following is simple algorithm for finding Preorder Predecessor of binary tree.

Step 1: Current root itself is NULL, then predecessor is also NULL.

Step 2: Current root contains value same as key for which we are looking predecessor.

        2.1: If current root is the root of the tree, then pred(current root) is undefined

        2.2: If u has a left sibling, ls, then pred(current root) is the rightmost descendant of ls
        2.3: Otherwise, pred(current root) is parent(current root).

Step 3: Current root is not the target node for which we are looking predecessor.

        3.1: Search target node and its predecessor in left side of tree recursively, and return if found.

        3.2: Search target node and its predecessor in right side of tree recursively, and return.

Following is implementation of above algorithm in c language.

Please write comments if you find anything wrong or you want to add something more related to this topic.

Algo#4: Postorder Successor in Binary Tree

Postorder Successor value for any node X means value of node Y that comes just after node X while doing Postorder traversal. For given tree in figure, postorder traversal (Left Right Root) would be : 4 5 2 6 3 1. We can search postorder successor for any node by just looking at this traversal output. E.G. postorder successor of node '4' is node '5' because '5' appears after '4' in given traversal.

But in real life we encounter Postorder Successor to be search for randomly any node, and if we go by this method of doing postorder traversal first then giving successor  it would be inefficient because of its time complexity of O(n) in large trees. So we need some method which can return Postorder successor randomly for any node without doing traversal explicitly in less time complexity.

Luckily postorder traversal follows some pattern which can help us in giving Postorder Successor for any node without doing traversal at all.

Following is simple algorithm for finding Postorder Successor of binary tree.

Step 1: If current root is NULL, then succ(current root) is NULL.

Step 2: If current root is target node for which we are looking for successor.

        2.1: If current root is the root of the tree, succ(current root) is undefined.

        2.2: Otherwise, if current root is a right child, succ(current root) is parent(current root).

        2.3: Otherwise current root is a left child and the following applies:

           2.3.1: If u has a right sibling, r, succ(current root) is the leftmost leaf in r's sub-tree

           2.3.2: Otherwise succ(current root) is parent(current root).
           2.3.3: If none of above applies, then succ(current root) doesn't exist.

Step 3: Current root is not the target node for which we are looking predecessor.

        3.1: Search target node and its predecessor in left side of tree recursively, and return if found.

        3.2: Search target node and its predecessor in right side of tree recursively, and return.

Following is implementation of above algorithm in c language.

/*
Algo#4: Postorder Successor in Binary Tree

Description: Find Postorder Successor value in given Biinary Tree.
*/
#include 
#include 

typedef struct node
{
    int data;
    struct node *left,*right;
} Node;

Node *leftmostLEAFnotNODE(Node *root)
{
    while (root->left || root->right)
    {
        if(root->left) root=root->left;
        else root=root->right;
    }

    return root;
}

Node *postorderSuccessorRec(Node *root, int key,Node *direct_parent, Node *parent,Node * treeroot)
{
    //Case 1: If current root is NULL, then succ(current root) is NULL.
    if (root==NULL)
        return 0;
    //Case 2: If current root is target node for which we are looking for successor.
    if (root->data == key)
    {
        //Case 2.1: If current root is the root of the tree, succ(current root) is undefined.
        if(root == treeroot)
            return NULL;
        //Case 2.2: Otherwise, if current root is a right child, succ(current root) is parent(current root).
        else if(direct_parent != NULL && direct_parent->right==root)
            return direct_parent;
        //Case 2.3: Otherwise current root is a left child and the following applies:

        //Case 2.3.1: If u has a right sibling, r, succ(current root) is the leftmost leaf in r's sub-tree
        else if(direct_parent != NULL && direct_parent->left==root && direct_parent->right!=NULL)
            return leftmostLEAFnotNODE(direct_parent->right);
        //Case 2.3.2: Otherwise succ(current root) is parent(current root).
        else if(direct_parent != NULL && direct_parent->left==root && direct_parent->right==NULL)
            return direct_parent;
        //Case 2.3.3: If none of above applies, then succ(current root) doesn't exist.
        else
            return NULL;
    }
    //Case 3: Current root is not the target node for which we are looking successor.
    else
    {
        //Case 3.1: Search target node and its successor in left side of tree recursively, and return if found.
        Node *left=postorderSuccessorRec(root->left,key,root,root,treeroot);
        if (left)
            return left;
        //Case 3.2: Search target node and its successor in right side of tree recursively, and return.
        return postorderSuccessorRec(root->right,key,root,parent,treeroot);
    }
}

Node *postorderSuccessor(Node *root, int key)
{
    return postorderSuccessorRec(root,key,NULL,NULL,root);
}

struct node* newNode(int data)
{
  struct node* node = (struct node*)  malloc(sizeof(struct node));
  node->data = data;
  node->left = NULL;
  node->right = NULL;

  return(node);
}


int main()
{

  /*
            1
          /   \
        2      3
      /  \    /
    4     5  6
  */
  struct node *root = newNode(1);
  root->left        = newNode(2);
  root->right       = newNode(3);
  root->left->left  = newNode(4);
  root->left->right = newNode(5);
  root->right->left = newNode(6);

  struct node * succ;
  succ = postorderSuccessor(root, 3);
  printf("Postorder Successor of %d is : %d\n",3,succ?succ->data:0);

  succ = postorderSuccessor(root, 4);
  printf("Postorder Successor of %d is : %d\n",4,succ?succ->data:0);

  getchar();
  return 0;
}

Please write comments if you find anything wrong or you want to add something more related to this topic.

Algo#3: Postorder Predecessor in Binary Tree

Postorder Predecessor value for any node X means value of node Y that comes just before node X while doing Postorder traversal. For given tree in figure, postorder traversal (Left Right Root) would be : 4 5 2 6 3 1. We can search postorder predecessor for any node by just looking at this traversal output. E.G. postorder predecessor of node '5' is node '4' because '4' appears before '5' in given traversal.

But in real life we encounter Postorder Predecessor to be search for randomly any node, and if we go by this method of doing postorder traversal first then giving predecessor, it would be inefficient because of its time complexity of O(n) in large trees. So we need some method which can return Postorder Predecessor randomly for any node without doing traversal explicitly in less time complexity.

Luckily postorder traversal follows some pattern which can help us in giving Postorder Predecessor for any node without doing traversal at all.

Following is simple algorithm for finding Postorder Predecessor of binary tree.

Step 1: If current root is NULL, then pred(current root) is NULL.

Step 2: If current root is target node for which we are looking for predecessor.

        2.1: If current root has a right child, r, then pred(current root) is r.

        2.2: Otherwise If current root has a left child, l, then pred(current root) is l.

        2.3: Otherwise if current root has a left sibling, ls, then pred(current root) is ls

      2.4: Otherwise if current root has an ancestor, v, which is a right child and has a left sibling, vls, then pred(current root) is vls

        2.5: Otherwise, pred(current root) is undefined.

Step 3: Current root is not the target node for which we are looking predecessor.

        3.1: Search target node and its predecessor in left side of tree recursively, and return if found.

        3.2: Search target node and its predecessor in right side of tree recursively, and return.

Following is implementation of above algorithm in c language.

Please write comments if you find anything wrong or you want to add something more related to this topic.

Algo#2: Inorder Successor in Binary Tree

Inorder Successor value for any node X means value of node Y that comes just after node X while doing Inorder traversal. For given tree in figure, inorder traversal (Left Root Right) would be : 4 2 5 1 6 3. We can search inorder successor for any node by just looking at this traversal output. E.G. inorder successor of node '1' is node '6' because '6' appears after '1' in given traversal.

But in real life we encounter Inorder Successor to be search for any node randomly, and if we go by this method of doing inorder traversal first then giving successor, it would be inefficient because of its time complexity of O(n) in large trees. So we need some method which can return Inorder Successor randomly for any node without doing traversal explicitly in less time complexity.

Luckily inorder traversal follows some pattern which can help us in giving Inorder Successor for any node without doing traversal at all.

Following is simple algorithm for finding Inorder Successor of binary tree.

Step 1: Current root itself is NULL, then successor is also NULL.

Step 2: Current root contains value same as key for which we are looking successor.

        2.1: Current root has right child, then left most node of right child is successor.

        2.2: Current root doesn't has right child, then parent of current root is successor.

Step 3: Current root is not the target node for which we are looking successor.

        3.1: Search target node and its successor in left side of tree recursively, and return if found.

        3.2: Search target node and its successor in right side of tree recursively, and return.

Following is implementation of above algorithm in c language.

/*
Algo#2: Inorder Successor in Binary Tree

Description: Find Inorder Successor value in given Binary Tree.
*/
#include 
#include 

typedef struct node
{
    int data;
    struct node *left,*right;
} Node;

Node *findLeftMostNode(Node *root)
{
    while (root->left)
        root=root->left;
    return root;
}

Node *inorderSuccessorRec(Node *root, int key, Node *parent)
{
    //Case 1: Current root itself is NULL, then successor is also NULL.
    if (root==NULL)
        return 0;
    //Case 2: Current root contains value same as key for which we are looking successor.
    if (root->data == key)
    {
        //Case 2.1: Current root has right child, then left most node of right child is successor.
        if (root->right)
            return findLeftMostNode(root->right);
        //Case 2.2: Current root doesn't has right child, then parent of current root is successor.
        else
            return parent;
    }
    //Case 3: Current root is not the target node for which we are looking successor.
    else
    {
        //Case 3.1: Search target node and its successor in left side of tree recursively, and return if found.
        Node *left=inorderSuccessorRec(root->left,key,root);
        if (left)
            return left;
        //Case 3.2: Search target node and its successor in right side of tree recursively, and return.
        return inorderSuccessorRec(root->right,key,parent);
    }
}

Node *inorderSuccessor(Node *root, int key)
{
    return inorderSuccessorRec(root,key,NULL);
}

struct node* newNode(int data)
{
  struct node* node = (struct node*)  malloc(sizeof(struct node));
  node->data = data;
  node->left = NULL;
  node->right = NULL;

  return(node);
}

int main()
{

  /*
            1
          /   \
        2      3
      /  \    /
    4     5  6
  */
  struct node *root = newNode(1);
  root->left        = newNode(2);
  root->right       = newNode(3);
  root->left->left  = newNode(4);
  root->left->right = newNode(5);
  root->right->left = newNode(6);

  struct node * succ;

  succ = inorderSuccessor(root, 1);
  printf("Inorder Successor of %d is : %d\n",1,succ->data);

  succ = inorderSuccessor(root, 2);
  printf("Inorder Successor of %d is : %d\n",2,succ->data);

  getchar();
  return 0;
}

Please write comments if you find anything wrong or you want to add something more related to this topic.

Algo#1: Inorder Predecessor in Binary Tree

Inorder Predecessor value for any node X means value of node Y that comes just before node X while doing Inorder traversal. For given tree in figure, inorder traversal (Left Root Right) would be : 4 2 5 1 6 3. We can search inorder predecessor for any node by just looking at this traversal output. E.G. inorder predecessor of node '1' is node '5' because '5' appears before '1' in given traversal.

But in real life we encounter Inorder Predecessor to be search for randomly any node, and if we go by this method of doing inorder traversal first then giving predecessor, it would be inefficient because of its time complexity of O(n) in large trees. So we need some method which can return Inorder Predecessor randomly for any node without doing traversal explicitly in less time complexity.

Luckily inorder traversal follows some pattern which can help us in giving Inorder Predecessor for any node without doing traversal at all.

Following is simple algorithm for finding Inorder Predecessor of binary tree.

Step 1: Current root itself is NULL, then predecessor is also NULL.

Step 2: Current root contains value same as key for which we are looking predecessor.

        2.1: Current root has left child, then right most node of left child is predecessor.

        2.2: Current root doesn't has left child, then parent of current root is predecessor.

Step 3: Current root is not the target node for which we are looking predecessor.

        3.1: Search target node and its predecessor in left side of tree recursively, and return if found.

        3.2: Search target node and its predecessor in right side of tree recursively, and return.

Following is implementation of above algorithm in c language.

/*
Algo#1: Inorder Predecessor in Binary Tree

Description: Find Inorder Predecessor value in given Binary Tree.
*/

#include 
#include 

typedef struct node
{
    int data;
    struct node *left,*right;
} Node;

Node *findRightMostNode(Node *root)
{
    while (root->right)
        root=root->right;
    return root;
}

Node *inorderPredecessorRec(Node *root, int key, Node *parent)
{
    //Case 1: Current root  itself is NULL, then predecessor is also NULL.
    if (root==NULL)
        return 0;
    //Case 2: Current root  contains value same as key for which we are looking predecessor.
    if (root->data == key)
    {
        //Case 2.1: Current root has left child, then right most node of left child is predecessor.
        if (root->left)
            return findRightMostNode(root->left);
        //Case 2.2: Current root  doesn't has left child, then parent of current root is predecessor.
        else
            return parent;
    }
    //Case 3: Current root  is not the target node for which we are looking predecessor.
    else
    {
     //Case 3.1: Search target node and its predecessor in left side of tree recursively, and return if found.
        Node *left=inorderPredecessorRec(root->left,key,parent);
        if (left)
            return left;
        //Case 3.2: Search target node and its predecessor in right side of tree recursively, and return.
        return inorderPredecessorRec(root->right,key,root);
    }
}

Node *inorderPredecessor(Node *root, int key)
{
    return inorderPredecessorRec(root,key,NULL);
}

struct node* newNode(int data)
{
  struct node* node = (struct node*)
                       malloc(sizeof(struct node));
  node->data = data;
  node->left = NULL;
  node->right = NULL;

  return(node);
}

void deleteTree(struct node* node) 
{
    if (node == NULL) return;
 
    deleteTree(node->left);
    deleteTree(node->right);
   
    printf("\n Delete node: %d", node->data);
    free(node);
} 

int main()
{
  /*
            1
          /   \
        2      3
      /  \    /
    4     5  8
  */
  struct node *root = newNode(1);
  root->left        = newNode(2);
  root->right       = newNode(3);
  root->left->left  = newNode(4);
  root->left->right = newNode(5);
  root->right->left = newNode(8);

  struct node * prede;
  prede = inorderPredecessor(root, 1);
  printf("Inorder Predecessor of %d is : %d\n",1,prede?prede->data:0);

  prede = inorderPredecessor(root, 2);
  printf("Inorder Predecessor of %d is : %d\n",2,prede?prede->data:0);

  deleteTree(root);
  getchar();
  return 0;
}

Please write comments if you find anything wrong or you want to add something more related to this topic.