From 4a07830561748f1d1e0319f6bc9837deb7d7eace Mon Sep 17 00:00:00 2001
From: Shen Jie <drfish.me@gmail.com>
Date: Wed, 7 Jun 2017 23:47:16 +0800
Subject: [PATCH] Java solution 102

---
 java/_102BinaryTreeLevelOrderTraversal.java | 52 +++++++++++++++++++++
 1 file changed, 52 insertions(+)
 create mode 100644 java/_102BinaryTreeLevelOrderTraversal.java

diff --git a/java/_102BinaryTreeLevelOrderTraversal.java b/java/_102BinaryTreeLevelOrderTraversal.java
new file mode 100644
index 0000000..0d744c7
--- /dev/null
+++ b/java/_102BinaryTreeLevelOrderTraversal.java
@@ -0,0 +1,52 @@
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ * Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
+ * <p>
+ * For example:
+ * Given binary tree [3,9,20,null,null,15,7],
+ *   3
+ *  / \
+ * 9  20
+ *   /  \
+ * 15   7
+ * <p>
+ * return its level order traversal as:
+ * [
+ *   [3],
+ *   [9,20],
+ *   [15,7]
+ * ]
+ * <p>
+ * Created by drfish on 29/05/2017.
+ */
+public class _102BinaryTreeLevelOrderTraversal {
+    public List<List<Integer>> levelOrder(TreeNode root) {
+        List<List<Integer>> result = new ArrayList<>();
+        List<TreeNode> curr = new ArrayList<>();
+
+
+        if (root == null) {
+            return result;
+        }
+        curr.add(root);
+
+        while (!curr.isEmpty()) {
+            List<TreeNode> next = new ArrayList<>();
+            List<Integer> level = new ArrayList<>();
+            for (TreeNode node : curr) {
+                if (node.left != null) {
+                    next.add(node.left);
+                }
+                if (node.right != null) {
+                    next.add(node.right);
+                }
+                level.add(node.val);
+            }
+            result.add(level);
+            curr = next;
+        }
+        return result;
+    }
+}