n-ary-tree-level-order-traversal (N 叉树的层序遍历)

Author: w2xi

README.md

@@ -68,6 +68,7 @@
 |350|[两个数组的交集 II](https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/)|[JavaScript](./algorithms/intersection-of-two-arrays-ii.js)|Easy|
 |387|[字符串中的第一个唯一字符](https://leetcode-cn.com/problems/first-unique-character-in-a-string/)|[JavaScript](./algorithms/first-unique-character-in-a-string.js)|Easy|
 |404|[左叶子之和](https://leetcode.cn/problems/sum-of-left-leaves/)|[JavaScript](./algorithms/sum-of-left-leaves.js)|Easy|
+|429|[N 叉树的层序遍历](https://leetcode.cn/problems/n-ary-tree-level-order-traversal/)|[JavaScript](./algorithms/n-ary-tree-level-order-traversal.js)|Medium|
 |437|[路径总和 III](https://leetcode.cn/problems/path-sum-iii/)|[JavaScript](./algorithms/path-sum-iii.js)|Medium|
 |509|[斐波那契数](https://leetcode.cn/problems/fibonacci-number/)|[JavaScript](./algorithms/fibonacci-number.js)|Easy|
 |543|[二叉树的直径](https://leetcode.cn/problems/diameter-of-binary-tree/)|[JavaScript](./algorithms/diameter-of-binary-tree.js)|Easy|

algorithms/n-ary-tree-level-order-traversal.js

@@ -0,0 +1,58 @@
+/**
+ * // Definition for a Node.
+ * function Node(val,children) {
+ *    this.val = val;
+ *    this.children = children;
+ * };
+ */
+
+/**
+ * N 叉树的层序遍历
+ * @param {Node|null} root
+ * @return {number[][]}
+ */
+var levelOrder = function (root) {
+  // 1. DFS
+  const result = [];
+  const dfs = (node, level = 0) => {
+    if (node) {
+      if (!result[level]) result[level] = [];
+      result[level].push(node.val);
+      const children = node.children;
+      children.forEach((child) => {
+        dfs(child, level + 1);
+      });
+    }
+  };
+  dfs(root);
+
+  return result;
+
+  // BFS
+  // return bfs(root);
+};
+
+function bfs(root) {
+  const result = [];
+  const queue = [];
+  queue.push(root);
+
+  if (!root) {
+    return [];
+  }
+
+  while (queue.length > 0) {
+    const temp = [];
+    let size = queue.length;
+
+    while (size--) {
+      const node = queue.shift();
+      temp.push(node.val);
+      const children = node.children;
+      children.forEach((child) => queue.push(child));
+    }
+    result.push(temp);
+  }
+
+  return result;
+}