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 the implementation of the idea.
C++
// 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; } |
Python3
# Python3 program to sink odd nodes # to the bottom of binary tree # A binary tree node # Helper function to allocates a new node class newnode: # Constructor to create a new node def __init__(self, key): self.data = key self.left = None self.right = None # Helper function to check # if node is leaf node def isLeaf(root): return (root.left == None and root.right == None) # 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 def sink(root): # If None or is a leaf, do nothing if (root == None or isLeaf(root)): return # if left subtree exists and # left child is even if (root.left and not(root.left.data & 1)): # swap root's data with left child # and fix left subtree root.data, \ root.left.data = root.left.data, \ root.data sink(root.left) # if right subtree exists and # right child is even elif(root.right and not(root.right.data & 1)): # swap root's data with right child # and fix right subtree root.data, \ root.right.data = root.right.data, \ root.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 def sinkOddNodes(root): # If None or is a leaf, do nothing if (root == None or 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. def printLevelOrder(root): q = [] q.append(root) # Do Level order traversal while (len(q)): nodeCount = len(q) # Print one level at a time while (nodeCount): node = q[0] print(node.data, end = " ") q.pop(0) if (node.left != None): q.append(node.left) if (node.right != None): q.append(node.right) nodeCount -= 1 # Line separator for levels print() # Driver Code """ Constructed binary tree is 1 / \ 5 8 / \ / \ 2 4 9 10 """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) print("Level order traversal of modified tree") printLevelOrder(root) # This code is contributed by SHUBHAMSINGH10 |
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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