# Queries to find distance between two nodes of a Binary tree – O(logn) method

Given a binary tree, the task is to find the distance between two keys in a binary tree, no parent pointers are given. Distance between two nodes is the minimum number of edges to be traversed to reach one node from other.
This problem has been already discussed in previous post but it uses three traversals of the Binary tree, one for finding Lowest Common Ancestor(LCA) of two nodes(let A and B) and then two traversals for finding distance between LCA and A and LCA and B which has O(n) time complexity. In this post, a method will be discussed that requires the O(log(n)) time to find LCA of two nodes. ## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The distance between two nodes can be obtained in terms of lowest common ancestor. Following is the formula.

```Dist(n1, n2) = Dist(root, n1) + Dist(root, n2) - 2*Dist(root, lca)
'n1' and 'n2' are the two given keys
'root' is root of given Binary Tree.
'lca' is lowest common ancestor of n1 and n2
Dist(n1, n2) is the distance between n1 and n2.
```

Above formula can also be written as:

```Dist(n1, n2) = Level[n1] + Level[n2] - 2*Level[lca]
```

This problem can be breakdown into:

1. Finding levels of each node
2. Finding the Euler tour of binary tree
3. Building segment tree for LCA,

These steps are explained below :

1. Find the levels of each node by applying level order traversal.
2. Find the LCA of two nodes in binary tree in O(logn) by Storing Euler tour of Binary tree in array and computing two other arrays with the help of levels of each node and Euler tour.
These steps are shown below:

(I) First, find Euler Tour of binary tree. Euler tour of binary tree in example

(II) Then, store levels of each node in Euler array in a different array. (III) Then, store First occurrences of all nodes of binary tree in Euler array. H stores the indices of nodes from Euler array, so that range of query for finding minimum can be minimized and their by further optimizing the query time. 3. Then build segment tree on L array and take the low and high values from H array that will give us the first occurrences of say Two nodes(A and B) . Then, <strong>we query segment tree to find the minimum value say X in range (H[A] to H[B]). Then we use the index of value X as index to Euler array to get LCA, i.e. Euler[index(X)].

Let, A = 8 and B = 5.
(I) H = 1 and H =2
(II) Querying on Segment tree, we get min value in L array between 1 and 2 as X=0, index=7
(III) Then, LCA= Euler, i.e LCA = 1.

4. Finally, we apply distance formula discussed above to get distance between two nodes.

 `// C++ program to find distance between ` `// two nodes for multiple queries ` `#include ` `#define MAX 100001 ` `using` `namespace` `std; ` ` `  `/* A tree node structure */` `struct` `Node { ` `    ``int` `data; ` `    ``struct` `Node* left; ` `    ``struct` `Node* right; ` `}; ` ` `  `/* Utility function to create a new Binary Tree node */` `struct` `Node* newNode(``int` `data) ` `{ ` `    ``struct` `Node* temp = ``new` `struct` `Node; ` `    ``temp->data = data; ` `    ``temp->left = temp->right = NULL; ` `    ``return` `temp; ` `} ` ` `  `// Array to store level of each node ` `int` `level[MAX]; ` ` `  `// Utility Function to store level of all nodes ` `void` `FindLevels(``struct` `Node* root) ` `{ ` `    ``if` `(!root) ` `        ``return``; ` ` `  `    ``// queue to hold tree node with level ` `    ``queue > q; ` ` `  `    ``// let root node be at level 0 ` `    ``q.push({ root, 0 }); ` ` `  `    ``pair<``struct` `Node*, ``int``> p; ` ` `  `    ``// Do level Order Traversal of tree ` `    ``while` `(!q.empty()) { ` `        ``p = q.front(); ` `        ``q.pop(); ` ` `  `        ``// Node p.first is on level p.second ` `        ``level[p.first->data] = p.second; ` ` `  `        ``// If left child exits, put it in queue ` `        ``// with current_level +1 ` `        ``if` `(p.first->left) ` `            ``q.push({ p.first->left, p.second + 1 }); ` ` `  `        ``// If right child exists, put it in queue ` `        ``// with current_level +1 ` `        ``if` `(p.first->right) ` `            ``q.push({ p.first->right, p.second + 1 }); ` `    ``} ` `} ` ` `  `// Stores Euler Tour ` `int` `Euler[MAX]; ` ` `  `// index in Euler array ` `int` `idx = 0; ` ` `  `// Find Euler Tour ` `void` `eulerTree(``struct` `Node* root) ` `{ ` ` `  `    ``// store current node's data ` `    ``Euler[++idx] = root->data; ` ` `  `    ``// If left node exists ` `    ``if` `(root->left) { ` ` `  `        ``// traverse left subtree ` `        ``eulerTree(root->left); ` ` `  `        ``// store parent node's data ` `        ``Euler[++idx] = root->data; ` `    ``} ` ` `  `    ``// If right node exists ` `    ``if` `(root->right) { ` `        ``// traverse right subtree ` `        ``eulerTree(root->right); ` ` `  `        ``// store parent node's data ` `        ``Euler[++idx] = root->data; ` `    ``} ` `} ` ` `  `// checks for visited nodes ` `int` `vis[MAX]; ` ` `  `// Stores level of Euler Tour ` `int` `L[MAX]; ` ` `  `// Stores indices of first occurrence ` `// of nodes in Euler tour ` `int` `H[MAX]; ` ` `  `// Preprocessing Euler Tour for finding LCA ` `void` `preprocessEuler(``int` `size) ` `{ ` `    ``for` `(``int` `i = 1; i <= size; i++) { ` `        ``L[i] = level[Euler[i]]; ` ` `  `        ``// If node is not visited before ` `        ``if` `(vis[Euler[i]] == 0) { ` `            ``// Add to first occurrence ` `            ``H[Euler[i]] = i; ` ` `  `            ``// Mark it visited ` `            ``vis[Euler[i]] = 1; ` `        ``} ` `    ``} ` `} ` ` `  `// Stores values and positions ` `pair<``int``, ``int``> seg[4 * MAX]; ` ` `  `// Utility function to find minimum of ` `// pair type values ` `pair<``int``, ``int``> min(pair<``int``, ``int``> a, ` `                   ``pair<``int``, ``int``> b) ` `{ ` `    ``if` `(a.first <= b.first) ` `        ``return` `a; ` `    ``else` `        ``return` `b; ` `} ` ` `  `// Utility function to build segment tree ` `pair<``int``, ``int``> buildSegTree(``int` `low, ``int` `high, ``int` `pos) ` `{ ` `    ``if` `(low == high) { ` `        ``seg[pos].first = L[low]; ` `        ``seg[pos].second = low; ` `        ``return` `seg[pos]; ` `    ``} ` `    ``int` `mid = low + (high - low) / 2; ` `    ``buildSegTree(low, mid, 2 * pos); ` `    ``buildSegTree(mid + 1, high, 2 * pos + 1); ` ` `  `    ``seg[pos] = min(seg[2 * pos], seg[2 * pos + 1]); ` `} ` ` `  `// Utility function to find LCA ` `pair<``int``, ``int``> LCA(``int` `qlow, ``int` `qhigh, ``int` `low, ` `                   ``int` `high, ``int` `pos) ` `{ ` `    ``if` `(qlow <= low && qhigh >= high) ` `        ``return` `seg[pos]; ` ` `  `    ``if` `(qlow > high || qhigh < low) ` `        ``return` `{ INT_MAX, 0 }; ` ` `  `    ``int` `mid = low + (high - low) / 2; ` ` `  `    ``return` `min(LCA(qlow, qhigh, low, mid, 2 * pos), ` `               ``LCA(qlow, qhigh, mid + 1, high, 2 * pos + 1)); ` `} ` ` `  `// Function to return distance between ` `// two nodes n1 and n2 ` `int` `findDistance(``int` `n1, ``int` `n2, ``int` `size) ` `{ ` `    ``// Maintain original Values ` `    ``int` `prevn1 = n1, prevn2 = n2; ` ` `  `    ``// Get First Occurrence of n1 ` `    ``n1 = H[n1]; ` ` `  `    ``// Get First Occurrence of n2 ` `    ``n2 = H[n2]; ` ` `  `    ``// Swap if low > high ` `    ``if` `(n2 < n1) ` `        ``swap(n1, n2); ` ` `  `    ``// Get position of minimum value ` `    ``int` `lca = LCA(n1, n2, 1, size, 1).second; ` ` `  `    ``// Extract value out of Euler tour ` `    ``lca = Euler[lca]; ` ` `  `    ``// return calculated distance ` `    ``return` `level[prevn1] + level[prevn2] - 2 * level[lca]; ` `} ` ` `  `void` `preProcessing(Node* root, ``int` `N) ` `{ ` `    ``// Build Tree ` `    ``eulerTree(root); ` ` `  `    ``// Store Levels ` `    ``FindLevels(root); ` ` `  `    ``// Find L and H array ` `    ``preprocessEuler(2 * N - 1); ` ` `  `    ``// Build segment Tree ` `    ``buildSegTree(1, 2 * N - 1, 1); ` `} ` ` `  `/* Driver function to test above functions */` `int` `main() ` `{ ` `    ``int` `N = 8; ``// Number of nodes ` ` `  `    ``/* Constructing tree given in the above figure */` `    ``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); ` `    ``root->right->right = newNode(7); ` `    ``root->right->left->right = newNode(8); ` ` `  `    ``// Function to do all preprocessing ` `    ``preProcessing(root, N); ` ` `  `    ``cout << ``"Dist(4, 5) = "` `<< findDistance(4, 5, 2 * N - 1) << ``"\n"``; ` `    ``cout << ``"Dist(4, 6) = "` `<< findDistance(4, 6, 2 * N - 1) << ``"\n"``; ` `    ``cout << ``"Dist(3, 4) = "` `<< findDistance(3, 4, 2 * N - 1) << ``"\n"``; ` `    ``cout << ``"Dist(2, 4) = "` `<< findDistance(2, 4, 2 * N - 1) << ``"\n"``; ` `    ``cout << ``"Dist(8, 5) = "` `<< findDistance(8, 5, 2 * N - 1) << ``"\n"``; ` ` `  `    ``return` `0; ` `} `

Output:

```Dist(4, 5) = 2
Dist(4, 6) = 4
Dist(3, 4) = 3
Dist(2, 4) = 1
Dist(8, 5) = 5
```

Time Complexity: O(Log N)
Space Complexity: O(N)

Queries to find distance between two nodes of a Binary tree – O(1) method

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