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题目：
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,#,#,15,7},
3
/ 9 20
/ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
题解：
这道题就是用传统的BFS来做。代码如下：
1 public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) { 2 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>(); 3 if(root == null) 4 return res; 5 6 LinkedList<TreeNode> queue = new LinkedList<TreeNode>(); 7 queue.add(root); 8 int curLevCnt = 1; 9 int nextLevCnt = 0; 10 ArrayList<Integer> levelres = new ArrayList<Integer>(); 11 12 while(!queue.isEmpty()){ 13 TreeNode cur = queue.poll(); 14 curLevCnt--; 15 levelres.add(cur.val); 16 17 if(cur.left != null){ 18 queue.add(cur.left); 19 nextLevCnt++; 20 } 21 if(cur.right != null){ 22 queue.add(cur.right); 23 nextLevCnt++; 24 } 25 26 if(curLevCnt == 0){ 27 curLevCnt = nextLevCnt; 28 nextLevCnt = 0; 29 res.add(levelres); 30 levelres = new ArrayList<Integer>(); 31 } 32 } 33 return res; 34 }
reference:http://www.cnblogs.com/springfor/p/3891391.html
*Binary Tree Level Order Traversal
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原文地址：http://www.cnblogs.com/hygeia/p/4704027.html
