Binary Tree Level Order Traversal

[解题思路]

记录下一层元素个数，这样每次遍历时就知道何时结束，只需一个queue

用一个count来计数每一层的node的个数，

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        
        if(root == null)
            return result;
            
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        //用来计数每一层有多少个Node
        int nextLevelCount = 1;
        
        while(!queue.isEmpty()){
            int curLevel = nextLevelCount;
            nextLevelCount = 0;
            ArrayList<Integer> lvl = new ArrayList<Integer>();
            // 此循环就是根据nextLevelCount来从队列里拿出多少个Node
            for(int i = 0; i < curLevel; i++){
                TreeNode tmp = queue.poll();
                lvl.add(tmp.val);
                if(tmp.left != null){
                    queue.add(tmp.left);
                    nextLevelCount++;
                }
                if(tmp.right != null){
                    queue.add(tmp.right);
                    nextLevelCount++;
                }
            }
            result.add(lvl);
        }
        return result;
    }
}

用两个queue的方法

 1 public class Solution {
 2     public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
 3         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
 4         if(root == null)
 5             return res;
 6             
 7         Queue<TreeNode> curlvl = new LinkedList<TreeNode>();
 8         
 9         
10         curlvl.add(root);
11         
12         while(!curlvl.isEmpty()){
13             ArrayList<Integer>  output = new ArrayList<Integer>();
14             Queue<TreeNode> nextlvl = new LinkedList<TreeNode>();
15             while(!curlvl.isEmpty()){
16                 TreeNode n = curlvl.poll();
17                 output.add(n.val);
18                 if(n.left != null){
19                     nextlvl.add(n.left);
20                 }
21                 if(n.right != null){
22                     nextlvl.add(n.right);
23                 }
24             }
25             res.add(output);
26             curlvl = nextlvl;
27         }
28         return res;
29     }
30 }

ref:http://www.cnblogs.com/feiling/p/3258537.html