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

public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if(root == null)
            return res;
            
        Queue<TreeNode> curlvl = new LinkedList<TreeNode>();
        
        
        curlvl.add(root);
        
        while(!curlvl.isEmpty()){
            ArrayList<Integer>  output = new ArrayList<Integer>();
            Queue<TreeNode> nextlvl = new LinkedList<TreeNode>();
            while(!curlvl.isEmpty()){
                TreeNode n = curlvl.poll();
                output.add(n.val);
                if(n.left != null){
                    nextlvl.add(n.left);
                }
                if(n.right != null){
                    nextlvl.add(n.right);
                }
            }
            res.add(output);
            curlvl = nextlvl;
        }
        return res;
    }
}

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