feat(curriculum): restore seed + solution to Chinese (#40683)

* feat(tools): add seed/solution restore script

* chore(curriculum): remove empty sections' markers

* chore(curriculum): add seed + solution to Chinese

* chore: remove old formatter

* fix: update getChallenges

parse translated challenges separately, without reference to the source

* chore(curriculum): add dashedName to English

* chore(curriculum): add dashedName to Chinese

* refactor: remove unused challenge property 'name'

* fix: relax dashedName requirement

* fix: stray tag

Remove stray `pre` tag from challenge file.

Signed-off-by: nhcarrigan <nhcarrigan@gmail.com>

Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>

curriculum/challenges/chinese/10-coding-interview-prep/data-structures/breadth-first-search.md

@@ -3,6 +3,7 @@ id: 587d825c367417b2b2512c90
 title: 广度优先搜索
 challengeType: 1
 videoUrl: ''
+dashedName: breadth-first-search
 ---
 
 # --description--
@@ -81,5 +82,85 @@ assert(
 );
 ```
 
+# --seed--
+
+## --after-user-code--
+
+```js
+// Source: http://adripofjavascript.com/blog/drips/object-equality-in-javascript.html
+function isEquivalent(a, b) {
+    // Create arrays of property names
+    var aProps = Object.getOwnPropertyNames(a);
+    var bProps = Object.getOwnPropertyNames(b);
+    // If number of properties is different,
+    // objects are not equivalent
+    if (aProps.length != bProps.length) {
+        return false;
+    }
+    for (var i = 0; i < aProps.length; i++) {
+        var propName = aProps[i];
+        // If values of same property are not equal,
+        // objects are not equivalent
+        if (a[propName] !== b[propName]) {
+            return false;
+        }
+    }
+    // If we made it this far, objects
+    // are considered equivalent
+    return true;
+}
+```
+
+## --seed-contents--
+
+```js
+function bfs(graph, root) {
+  var nodesLen = {};
+
+  return nodesLen;
+};
+
+var exBFSGraph = [
+  [0, 1, 0, 0],
+  [1, 0, 1, 0],
+  [0, 1, 0, 1],
+  [0, 0, 1, 0]
+];
+console.log(bfs(exBFSGraph, 3));
+```
+
 # --solutions--
 
+```js
+function bfs(graph, root) {
+  var nodesLen = {};
+  // Set all distances to infinity
+  for (var i = 0; i < graph.length; i++) {
+    nodesLen[i] = Infinity;
+  }
+  nodesLen[root] = 0; // ...except root node
+  var queue = [root]; // Keep track of nodes to visit
+  var current; // Current node traversing
+  // Keep on going until no more nodes to traverse
+  while (queue.length !== 0) {
+    current = queue.shift();
+    // Get adjacent nodes from current node
+    var curConnected = graph[current]; // Get layer of edges from current
+    var neighborIdx = []; // List of nodes with edges
+    var idx = curConnected.indexOf(1); // Get first edge connection
+    while (idx !== -1) {
+      neighborIdx.push(idx); // Add to list of neighbors
+      idx = curConnected.indexOf(1, idx + 1); // Keep on searching
+    }
+    // Loop through neighbors and get lengths
+    for (var j = 0; j < neighborIdx.length; j++) {
+      // Increment distance for nodes traversed
+      if (nodesLen[neighborIdx[j]] === Infinity) {
+        nodesLen[neighborIdx[j]] = nodesLen[current] + 1;
+        queue.push(neighborIdx[j]); // Add new neighbors to queue
+      }
+    }
+  }
+  return nodesLen;
+}
+```