Sink Odd nodes in Binary Tree

Given a Binary Tree having odd and even elements, sink all its odd valued nodes such that no node with odd value could be parent of node with even value. There can be multiple outputs for a given tree, we need to print one of them. It is always possible to convert a tree (Note that a node with even nodes and all odd nodes follows the rule)

Input : 
       1
    /    \
   2      3
Output
       2            2
    /    \   OR   /   \
   1      3      3     1 
  

Input : 
       1
     /    \
    5       8
  /  \     /  \
 2    4   9    10
Output :
    2                 4
  /    \            /    \     
 4       8    OR   2      8    OR .. (any tree with 
/  \    /  \      /  \   / \          same keys and 
5   1  9   10    5    1 9   10        no odd is parent
                                      of even)

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Basically, we need to swap odd value of a node with even value of one of its descendants. The idea is to traverse the tree in postorder fashion. Since we process in postorder, for each odd node encountered, its left and right subtrees are already balanced (sinked), we check if it’s an odd node and its left or right child has an even value. If even value is found, we swap the node’s data with that of even child node and call the procedure on the even child to balance the subtree. If both children have odd values, that means that all its descendants are odd.

Below is C++ implementation of the idea.

// Program to sink odd nodes to the bottom of 
// binary tree 
#include<bits/stdc++.h> 
using namespace std; 
  
// A binary tree node 
struct Node 
{ 
    int data; 
    Node* left, *right; 
}; 
  
// Helper function to allocates a new node 
Node* newnode(int data) 
{ 
    Node* node = new Node; 
    node->data = data; 
    node->left = node->right = NULL; 
    return node; 
} 
  
// Helper function to check if node is leaf node 
bool isLeaf(Node *root) 
{ 
    return (root->left == NULL && root->right == NULL); 
} 
  
// A recursive method to sink a tree with odd root 
// This method assumes that the subtrees are already 
// sinked. This method is similar to Heapify of 
// Heap-Sort 
void sink(Node *&root) 
{ 
    // If NULL or is a leaf, do nothing 
    if (root == NULL || isLeaf(root)) 
        return; 
  
    // if left subtree exists and left child is even 
    if (root->left && !(root->left->data & 1)) 
    { 
        // swap root's data with left child and 
        // fix left subtree 
        swap(root->data, root->left->data); 
        sink(root->left); 
    } 
  
    // if right subtree exists and right child is even 
    else if(root->right && !(root->right->data & 1)) 
    { 
        // swap root's data with right child and 
        // fix right subtree 
        swap(root->data, root->right->data); 
        sink(root->right); 
    } 
} 
  
// Function to sink all odd nodes to the bottom of binary 
// tree. It does a postorder traversal and calls sink() 
// if any odd node is found 
void sinkOddNodes(Node* &root) 
{ 
    // If NULL or is a leaf, do nothing 
    if (root == NULL || isLeaf(root)) 
        return; 
  
    // Process left and right subtrees before this node 
    sinkOddNodes(root->left); 
    sinkOddNodes(root->right); 
  
    // If root is odd, sink it 
    if (root->data & 1) 
        sink(root); 
} 
  
// Helper function to do Level Order Traversal of 
// Binary Tree level by level. This function is used 
// here only for showing modified tree. 
void printLevelOrder(Node* root) 
{ 
    queue<Node*> q; 
    q.push(root); 
  
    // Do Level order traversal 
    while (!q.empty()) 
    { 
        int nodeCount = q.size(); 
  
        // Print one level at a time 
        while (nodeCount) 
        { 
            Node *node = q.front(); 
            printf("%d ", node->data); 
            q.pop(); 
            if (node->left != NULL) 
                q.push(node->left); 
            if (node->right != NULL) 
                q.push(node->right); 
            nodeCount--; 
        } 
  
        // Line separator for levels 
        printf("\n"); 
    } 
} 
  
// Driver program to test above functions 
int main() 
{ 
    /* Constructed binary tree is 
            1 
          /   \ 
         5      8 
        / \   /  \ 
       2   4 9   10     */
  
    Node *root = newnode(1); 
    root->left = newnode(5); 
    root->right    = newnode(8); 
    root->left->left = newnode(2); 
    root->left->right = newnode(4); 
    root->right->left = newnode(9); 
    root->right->right = newnode(10); 
  
    sinkOddNodes(root); 
  
    printf("Level order traversal of modified tree\n"); 
    printLevelOrder(root); 
  
    return 0; 
} 

Output :

Level order traversal of modified tree
2 
4 8 
5 1 9 10

This article is contributed by Aditya Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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