# Print updated levels of each node of a Complete Binary Tree based on difference in weights of subtrees

Given a complete binary tree with **N** **levels** numbered **[0, (N – 1 )]** from root to the lowest level in decreasing order and having **weights** numbered between **[1, 2 ^{N} – 1]** from the root to the last leaf node in the increasing order, the task for every node is to adjust the values of the levels of all nodes present in its left and right subtree based on the following condition:

- Increase the level of all the nodes of lighter subtree by the difference of their weights.
- Decrease the level of all the nodes of heavier subtree by the difference of their weights.

**Examples:**

Input:1 / \ 2 3

Output:0 0 -2

Explanation:

The initial levels of the nodes {1,2,3} are {0,-1,-1} respectively.

The root node remains unchanged.

The weight of left subtree is 2 and the weight of the right subtree is 3.

So, the left subtree goes up by (3 – 2) = 1 level to 0.

The right subtree goes down by 1 level to -2.

Input:

1 / \ 2 3 / \ / \ 4 5 6 7

Output:0 4 -6 4 2 -6 -8

Explanation:

The initial levels of the nodes {1,2,3,4,5,6,7} are {0,-1,-1,-2,-2,-2,-2} respectively.

The root node remains unchanged.

The weight of the left subtree {2,4,5} is 11.

The weight of the right subtree {3,6,7} is 16.

Hence all the nodes in left subtree move up by 5 while those in the right subtree moves down by 5.

Thus, the new levels of every node are:

Node 2: -1 + 5 = 4

Node 3: -1 – 5 = -6

Node 4,5: -2 + 5 = 3

Node 6,7: -2 – 5 = -7

Now, nodes 4,5 further based on the difference of their weights (5 -4 ) = 1.

Node 4: 3 + 1 = 4

Node 5: 3 – 1 = 2

Similarly, nodes 6,7 also get adjusted.

Node 6: -7 + 1 = -6

Node 7: -7 – 1 = -8

Hence the final adjusted levels of all the nodes are 0 4 -6 4 2 -6 -8.

**Approach:** In order to solve this problem, we calculate the weights of the left **(w_left)** and right **(w_right)** subtrees for every node and store their difference **K**. Once calculated, we recursively increase the value of the **level** of all the nodes of its lighter subtree by **K** and decrease that of its heavier subtree by **K** from their respective current values. Once computed for all nodes, we display the final values of **level** of every node.

Below code is the implementation of the above approach:

## C++

`// C++ Program to print updated levels ` `// of each node of a Complete Binary Tree ` `// based on difference in weights of subtrees ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Node for the given binary tree ` `struct` `node { ` ` ` ` ` `int` `weight; ` ` ` ` ` `// stores the weight of node ` ` ` `int` `level; ` ` ` ` ` `// stores the level of node ` ` ` `struct` `node* left; ` ` ` `struct` `node* right; ` ` ` ` ` `node(` `int` `w, ` `int` `l) ` ` ` `{ ` ` ` `this` `->weight = w; ` ` ` `this` `->level = l; ` ` ` `left = right = NULL; ` ` ` `} ` `}; ` ` ` `// Utility function to insert a node ` `// in a tree rooted at root ` `struct` `node* insert(` `struct` `node* root, ` ` ` `int` `n_weight, ` `int` `n_level, queue<node*>& q) ` `{ ` ` ` ` ` `struct` `node* n ` `= ` `new` `node(n_weight, n_level); ` ` ` ` ` `// if the tree is empty till now ` ` ` `// make node n the root ` ` ` `if` `(root == NULL) ` ` ` `root = n; ` ` ` ` ` `// If the frontmost node of ` ` ` `// queue has no left child ` ` ` `// make node n its left child ` ` ` `// the frontmost node still ` ` ` `// remains in the queue because ` ` ` `// its right child is null yet ` ` ` `else` `if` `(q.front()->left == NULL) { ` ` ` `q.front()->left = n; ` ` ` `} ` ` ` ` ` `// Make node n the right child of ` ` ` `// the frontmost node and remove ` ` ` `// the front node from queue ` ` ` `else` `{ ` ` ` `q.front()->right = n; ` ` ` `q.pop(); ` ` ` `} ` ` ` `// push the node n into queue ` ` ` `q.push(n); ` ` ` ` ` `return` `root; ` `} ` ` ` `// Function to create a complete binary tree ` `struct` `node* createTree(vector<` `int` `> weights, ` `vector<` `int` `> levels) ` `{ ` ` ` ` ` `// initialise the root node of tree ` ` ` `struct` `node* root = NULL; ` ` ` ` ` `// initialise a queue of nodes ` ` ` `queue<node*> q; ` ` ` ` ` `int` `n = weights.size(); ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `/* ` ` ` `keep inserting nodes with weight values ` ` ` `from the weights vector and level values ` ` ` `from the levels vector ` ` ` `*/` ` ` `root = insert(root, weights[i], ` ` ` `levels[i], q); ` ` ` `} ` ` ` `return` `root; ` `} ` ` ` `// Function to print final levels of nodes ` `void` `printNodeLevels(` `struct` `node* root) ` `{ ` ` ` ` ` `if` `(root == NULL) ` ` ` `return` `; ` ` ` ` ` `queue<node*> q; ` ` ` `q.push(root); ` ` ` ` ` `while` `(!q.empty()) { ` ` ` ` ` `cout << q.front()->level << ` `" "` `; ` ` ` ` ` `if` `(q.front()->left != NULL) ` ` ` `q.push(q.front()->left); ` ` ` `if` `(q.front()->right != NULL) ` ` ` `q.push(q.front()->right); ` ` ` `q.pop(); ` ` ` `} ` ` ` `cout << endl; ` `} ` ` ` `// Function to find the weight of subtree ` `int` `findWeight(` `struct` `node* root) ` `{ ` ` ` `// If the root node is null ` ` ` `// then weight of subtree will be 0 ` ` ` `if` `(root == NULL) ` ` ` `return` `0; ` ` ` ` ` `return` `root->weight + ` ` ` `findWeight(root->left) ` ` ` `+ findWeight(root->right); ` `} ` ` ` `// Function to compute new level ` `// of the nodes based on the ` `// difference of weight K ` `void` `changeLevels(` `struct` `node* root, ` `int` `k) ` `{ ` ` ` ` ` `if` `(root == NULL) ` ` ` `return` `; ` ` ` ` ` `// Change the level of root node ` ` ` `root->level = root->level + k; ` ` ` ` ` `// Recursively change the level of ` ` ` `// left and right subtree ` ` ` `changeLevels(root->left, k); ` ` ` `changeLevels(root->right, k); ` `} ` ` ` `// Function to calculate weight of ` `// the left and the right subtrees and ` `// adjust levels based on their difference ` `void` `adjustLevels(` `struct` `node* root) ` `{ ` ` ` ` ` `// No adjustment required ` ` ` `// when root is null ` ` ` `if` `(root == NULL) ` ` ` `return` `; ` ` ` ` ` `// Find weights of left ` ` ` `// and right subtrees ` ` ` `int` `w_left = findWeight(root->left); ` ` ` `int` `w_right = findWeight(root->right); ` ` ` ` ` `// find the difference between the ` ` ` `// weights of left and right subtree ` ` ` `int` `w_diff = w_left - w_right; ` ` ` ` ` `// Change the levels of nodes ` ` ` `// according to weight difference at root ` ` ` `changeLevels(root->left, -w_diff); ` ` ` `changeLevels(root->right, w_diff); ` ` ` ` ` `// Recursively adjust the levels ` ` ` `// for left and right subtrees ` ` ` `adjustLevels(root->left); ` ` ` `adjustLevels(root->right); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `// Number of levels ` ` ` `int` `N = 3; ` ` ` ` ` `// Number of nodes ` ` ` `int` `nodes = ` `pow` `(2, N) - 1; ` ` ` ` ` `vector<` `int` `> weights; ` ` ` `// Vector to store weights of tree nodes ` ` ` `for` `(` `int` `i = 1; i <= nodes; i++) { ` ` ` `weights.push_back(i); ` ` ` `} ` ` ` ` ` `vector<` `int` `> levels; ` ` ` `// Vector to store levels of every nodes ` ` ` `for` `(` `int` `i = 0; i < N; i++) { ` ` ` ` ` `// 2^i nodes are present at ith level ` ` ` `for` `(` `int` `j = 0; j < ` `pow` `(2, i); j++) { ` ` ` ` ` `// value of level becomes negative ` ` ` `// while going down the root ` ` ` `levels.push_back(-1 * i); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Create tree with the ` `// given weights and levels ` ` ` `struct` `node* root ` `= createTree(weights, levels); ` ` ` ` ` `// Adjust the levels ` ` ` `adjustLevels(root); ` ` ` ` ` `// Display the final levels ` ` ` `printNodeLevels(root); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

0 4 -6 4 2 -6 -8

## Recommended Posts:

- Print Binary Tree levels in sorted order | Set 3 (Tree given as array)
- Print all K-sum levels in a Binary Tree
- Print all Co-Prime Levels of a Binary Tree
- Print all Prime Levels of a Binary Tree
- Print Levels of all nodes in a Binary Tree
- Print all Palindromic Levels Of a Binary Tree
- Print all Exponential Levels of a Binary Tree
- Print all nodes between two given levels in Binary Tree
- Print Binary Tree levels in sorted order
- Print Binary Tree levels in sorted order | Set 2 (Using set)
- Print even positioned nodes of odd levels in level order of the given binary tree
- Print odd positioned nodes of odd levels in level order of the given binary tree
- Print odd positioned nodes of even levels in level order of the given binary tree
- Print even positioned nodes of even levels in level order of the given binary tree
- Print path from root to all nodes in a Complete Binary Tree
- Print a Binary Tree in Vertical Order | Set 2 (Map based Method)
- Count of subtrees in a Binary Tree having XOR value K
- Print cousins of a given node in Binary Tree
- Print Ancestors of a given node in Binary Tree
- Print path from root to a given node in a binary tree

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.