In computer science, binary trees are hierarchical data structures that consist of nodes. Each node can have two children at most - one on the left side and one on the right. These structures have a top-to-bottom order. Solving for distance between any two given nodes is a problem often seen with binary trees. We'll look at different ways to find this distance in JavaScript.
Problem Description: Given a binary tree and two node values, the task is to find the distance between the two nodes.
Consider the following binary tree:
Explanation: To find the distance between nodes 4 and 6, we first locate their common ancestor, which is node 2. From this ancestor, the path to node 4 is of length 1, and the path to node 6 is of length 3. Therefore, the total distance between nodes 4 and 6 is 4.
Output: The distance between nodes 4 and 6 in the binary tree is 4.
There are several approaches to find the distance between two nodes of a Binary tree using JavaScript which are as follows:
Using Brute Force
In the brute force approach, we traverse the binary tree recursively, calculating the distance from the root to both target nodes. We then sum up these distances to find the total distance between the nodes.
- In this approach, for each node in the binary tree, we find the distance to both target nodes.
- We start from the root and traverse the tree recursively.
- We calculate the distance to both target nodes for each node encountered using depth-first search.
- Finally, we sum up the distances from the root to both nodes to get the total distance.
Example: To demonstrate finding the distance between two nodes of a Binary tree using brute force in JavaScript.
class BinaryNode {
constructor(val)
{
this.val = val;
this.left = null;
this.right = null;
}
}
// Function to find distance between
// two nodes using brute force approach
function findDistanceBruteForce(root, p, q)
{
// Function to find distance
// from root to a given node
function findDepth(node, target)
{
if (!node) return 0;
if (node.val === target) return 0;
const left = findDepth(node.left, target);
const right = findDepth(node.right, target);
if (left !== -1) return left + 1;
if (right !== -1) return right + 1;
return -1;
}
const distanceP = findDepth(root, p);
const distanceQ = findDepth(root, q);
return distanceP + distanceQ;
}
// Construct the binary tree
const root = new BinaryNode(1);
root.left = new BinaryNode(2);
root.right = new BinaryNode(3);
root.left.left = new BinaryNode(4);
root.left.right = new BinaryNode(5);
root.right.right = new BinaryNode(6);
const node1 = 4;
const node2 = 6;
// Find distance between node1 and node2 using brute force approach
console.log("Distance between", node1,
"and", node2, "using brute force approach is",
findDistanceBruteForce(root, node1, node2));
Output:
Distance between 4 and 6 using brute force approach is 4Time Complexity: O(n^2)
Space Complexity: O(h)
Using Lowest Common Ancestor (LCA)
To optimize the brute force approach, we utilize the concept of the Lowest Common Ancestor (LCA). We first find the LCA of the two target nodes, then calculate the distances from the LCA to both nodes, and finally sum up these distances to get the total distance between the nodes.
- We can optimize the previous approach by first finding the Lowest Common Ancestor (LCA) of the two target nodes.
- Once we have the LCA, we can calculate the distances from the LCA to both target nodes using depth-first search.
- Finally, we sum up these distances to get the total distance between the target nodes.
Example: To demonsrtate finding the distance between two nodes of a Binary tree using JavaScript Lowest common ancestor.
class TreeNode {
constructor(val) {
this.val = val;
this.left = null;
this.right = null;
}
}
// Function to find Lowest
// Common Ancestor (LCA)
function findLCA(root, p, q)
{
if (!root || root
.val === p || root
.val === q) return root;
const left = findLCA(root.left, p, q);
const right = findLCA(root.right, p, q);
if (left && right) return root;
return left ? left : right;
}
// Function to find distance
// between two nodes using LCA
function findDistanceOptimized(root, p, q) {
const lca = findLCA(root, p, q);
// Function to find distance
// from root to a given node
function findDepth(node, target, depth) {
if (!node) return 0;
if (node.val === target) return depth;
const left = findDepth(node.left, target, depth + 1);
const right = findDepth(node.right, target, depth + 1);
return left || right;
}
const distanceP = findDepth(lca, p, 0);
const distanceQ = findDepth(lca, q, 0);
return distanceP + distanceQ;
}
// Construct the binary tree
const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);
root.right.right = new TreeNode(6);
const node1 = 4;
const node2 = 6;
// Find distance between node1 and node2
// using optimized approach with LCA
console.log("Distance between",
node1, "and", node2, "using optimized approach with LCA is",
findDistanceOptimized(root, node1, node2));
Output:
Distance between 4 and 6 using optimized approach with LCA is 4Time Complexity: O(n)
Space Complexity: O(h)
Using Single Traversal
In this approach, we further optimize by finding both the LCA and the distances from the LCA to both nodes in a single traversal of the binary tree. This reduces the time complexity compared to the previous approaches.
- We can further optimize by finding both the LCA and the distances from the LCA to both nodes in a single traversal of the binary tree.
- During the traversal, if we find both target nodes, we return their depths to their common ancestor.
- If only one node is found, we return its depth and continue the traversal.
- Once both nodes are found, we return their depths to their common ancestor and calculate the total distance.
Example: To demonsrtate finding the distance between two nodes of Binary tree using single traversal in JavaScript.
class TreeNode {
constructor(val) {
this.val = val;
this.left = null;
this.right = null;
}
}
// Function to find the distance
// between two nodes in a binary tree
function findDistance(root, p, q) {
// Helper function to find the
// distance from a node to the target
function findDepth(node, target, depth) {
if (!node) return 0;
if (node.val === target) return depth;
const left = findDepth(node.left, target, depth + 1);
const right = findDepth(node.right, target, depth + 1);
return left || right;
}
// Helper function to find the Lowest
// Common Ancestor (LCA) of two nodes
function findLCA(node, p, q) {
if (!node || node.val === p || node.val === q) return node;
const left = findLCA(node.left, p, q);
const right = findLCA(node.right, p, q);
if (left && right) return node;
return left ? left : right;
}
// Find the Lowest Common
// Ancestor (LCA) of the two nodes
const lca = findLCA(root, p, q);
// Calculate the distance
// from the LCA to each node
const distanceP = findDepth(lca, p, 0);
const distanceQ = findDepth(lca, q, 0);
// Total distance between the nodes
return distanceP + distanceQ;
}
// Construct the binary tree
const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);
root.right.right = new TreeNode(6);
const node1 = 4;
const node2 = 6;
// Find distance between node1 and node2
console.log("Distance between", node1,
"and", node2,
"using further optimized approach with single traversal is",
findDistance(root, node1, node2));
Output:
Distance between 4 and 6 using further optimized approach with single traversal is 4Time Complexity: O(n)
Space Complexity: O(h)