binary tree level order traversal

fishercoder1534 · fishercoder1534 · commit 6755383869de · 2016-08-12T13:05:32.000-07:00
diff --git a/EASY/src/easy/BinaryTreeLevelOrderTraversal.java b/EASY/src/easy/BinaryTreeLevelOrderTraversal.java
@@ -0,0 +1,56 @@
+package easy;
+
+import classes.TreeNode;
+
+import java.util.ArrayList;
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Queue;
+
+/**102. Binary Tree Level Order Traversal
+
+    Total Accepted: 117221
+    Total Submissions: 341219
+    Difficulty: Easy
+
+Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
+
+For example:
+Given binary tree [3,9,20,null,null,15,7],
+
+    3
+   / \
+  9  20
+    /  \
+   15   7
+
+return its level order traversal as:
+
+[
+  [3],
+  [9,20],
+  [15,7]
+]
+*/
+public class BinaryTreeLevelOrderTraversal {
+	//Cheers, easily made it AC'ed on the first submission now, practice does make perfect!
+	public List<List<Integer>> levelOrder(TreeNode root) {
+		List<List<Integer>> result = new ArrayList<List<Integer>>();
+		if(root == null) return result;
+		
+		Queue<TreeNode> q = new LinkedList<TreeNode>();
+		q.offer(root);
+		while(!q.isEmpty()){
+			List<Integer> thisLevel = new ArrayList<Integer>();
+			int qSize = q.size();
+			for(int i = 0; i < qSize; i++){
+				TreeNode curr = q.poll();
+				thisLevel.add(curr.val);
+				if(curr.left != null) q.offer(curr.left);
+				if(curr.right != null) q.offer(curr.right);
+			}
+			result.add(thisLevel);
+		}
+		return result;
+	}
+}
