Sink even nodes in Binary Tree
Given a Binary Tree having odd and even elements, sink all it’s even valued nodes such that no node with even value could be a parent of a node with odd 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 odd nodes and all even nodes follows the rule)
Examples:
Input:
1
/ \
5 8
/ \ / \
2 4 9 10
Output:
1
5 9
2 4 8 10
Level order traversal after
sinking all the nodes
Input:
4
/ \
2 1
Output:
4
2 1
Explanation:
In the first case
Given tree
4
/ \
2 1
There are two trees possible
1 1
/ \ OR / \
2 4 4 2
In the second example also,
Given tree
1
/ \
5 8
/ \ / \
2 4 9 10
There are more than one tree
that can satisfy the condition
1 1
/ \ / \
5 9 OR 5 9
/ \ / \ / \ / \
2 4 8 10 4 2 8 10
Approach:
- Basically, it is required to swap the even value of a node with odd value of one of its descendants.
- The idea is to traverse the tree in postorder fashion.
- Since we process in postorder, for each even node encountered, it’s left and right subtrees are already balanced (sinked).
- Check if it’s an even node and its left or right child has an odd value. If odd value is found, swap the node’s data with that of odd child node and call the procedure on the odd child to balance the subtree.
- If both children have even values, that means that all its descendants are even.
Below is the implementation of the idea:
C/C++
// Program to sink even 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 odd 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 odd 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 even // nodes to the bottom of // binary tree. It does a // postorder traversal and // calls sink() // if any even node is found void sinkevenNodes(Node*& root) { // If NULL or is a // leaf, do nothing if (root == NULL || isLeaf(root)) return; // Process left and right // subtrees before this node sinkevenNodes(root->left); sinkevenNodes(root->right); // If root is even, 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 the node has a left // child then push into queue if (node->left != NULL) q.push(node->left); // If the node has a right // child then push into queue if (node->right != NULL) q.push(node->right); nodeCount--; } // Line separator for levels printf("\n"); } } // Driver code 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); // Calling function to perform // sink operation sinkevenNodes(root); // Printing the updated tree // using level oerder traversal printLevelOrder(root); return 0; } |
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Python
# Python3 program to sink even 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 even 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 (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 (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 even nodes to # the bottom of binary tree. It does # a postorder traversal and calls sink() # if any even node is found def sinkevenNodes(root): # If None or is a leaf, do nothing if (root == None or isLeaf(root)): return # Process left and right subtrees # before this node sinkevenNodes(root.left) sinkevenNodes(root.right) # If root is even, sink it if not (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) sinkevenNodes(root) printLevelOrder(root) # This code is contributed by SHUBHAMSINGH10 |
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Output:
1 5 9 2 4 8 10
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