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September 16, 2019 06:57
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Non-recursive DFS and BFS algorithms
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/** | |
* Depth-first and Breadth-first graph traversals. | |
* | |
* In this diff we implement non-recursive algorithms for DFS, | |
* and BFS maintaining an explicit stack and a queue. | |
* | |
* by Dmitry Soshnikov <dmitry.soshnikov@gmail.com> | |
* MIT Style license | |
*/ | |
class Node { | |
constructor(name, childNodes) { | |
this.name = name; | |
this.childNodes = childNodes; | |
this.visited = false; | |
} | |
} | |
// Nodes. | |
let A = new Node('A'); | |
let B = new Node('B'); | |
let C = new Node('C'); | |
let D = new Node('D'); | |
let E = new Node('E'); | |
let F = new Node('F'); | |
let G = new Node('G'); | |
let H = new Node('H'); | |
let allNodes = [A, B, C, D, E, F, G, H]; | |
function resetNodes() { | |
allNodes.forEach(node => { | |
node.visited = false; | |
}); | |
} | |
resetNodes(); | |
// Graph. | |
A.childNodes = [B, D, G]; | |
B.childNodes = [E, F]; | |
C.childNodes = [F, H]; | |
D.childNodes = [A, F]; | |
E.childNodes = [B, G]; | |
F.childNodes = [B, C, D]; | |
G.childNodes = [A, E]; | |
H.childNodes = [C]; | |
// --------------------------------------------------------------- | |
// 1. DFS | |
// --------------------------------------------------------------- | |
// DFS maintains an explicit stack of working nodes. | |
let stack = []; | |
// Before the beginning, the stack should contain | |
// the start node. We start from A. | |
stack.push(A); | |
let output = []; | |
function DFS() { | |
stackLoop: while (stack.length) { | |
// The top of the stack is our working node. | |
let node = stack[stack.length - 1]; | |
// Visit the node if it's not visited yet. | |
if (!node.visited) { | |
node.visited = true; | |
output.push(node); | |
} | |
// Get next node to visit. | |
for (let n of node.childNodes) { | |
if (!n.visited) { | |
// Found the node, just go to the DFS | |
// loop for it, pushing it onto the stack. | |
stack.push(n); | |
continue stackLoop; | |
} | |
} | |
// If we reach here, all child nodes were visited, | |
// so just pop the node from the stack. | |
stack.pop(); | |
} | |
} | |
DFS(); | |
// DFS: [ 'A', 'B', 'E', 'G', 'F', 'C', 'H', 'D' ] | |
console.log('DFS:', output.map(n => n.name)); | |
// Exercise: Implement a recursive version (basically, using | |
// the call-stack for the purposes of the algorithm) | |
// --------------------------------------------------------------- | |
// 2. BFS | |
// --------------------------------------------------------------- | |
// Clear visited flag. | |
resetNodes(); | |
output = []; | |
// BFS maintains a queue of working nodes. | |
let queue = []; | |
// Enqueue the start node, A. | |
queue.unshift(A); | |
function BFS() { | |
queueLoop: while (queue.length) { | |
// Get the next queued node to work on. | |
let node = queue.shift(); | |
// Visit the node if it's not visited yet. | |
if (!node.visited) { | |
node.visited = true; | |
output.push(node); | |
} | |
// Visit all direct child nodes, and | |
// enqueue them. | |
for (let n of node.childNodes) { | |
if (!n.visited) { | |
n.visited = true; | |
output.push(n); | |
queue.unshift(n); | |
} | |
} | |
// Go to the next node from the queue | |
// on the following iteration. | |
} | |
} | |
BFS(); | |
// BFS: [ 'A', 'B', 'D', 'G', 'E', 'F', 'C', 'H' ] | |
console.log('BFS:', output.map(n => n.name)); |
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