leetcode -- Binary Tree Level Order Traversal

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]
]

[解题思路]

1.可以维护两个queue，当前层一个，下一层一个

2.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

/**
 * 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) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(root == null){
            return result;
        }
        
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        int index = 0;
        int nextLevel = 1;
        while(!queue.isEmpty()){
            int curLevel = nextLevel;
            nextLevel = 0;
            ArrayList<Integer> lvl = new ArrayList<Integer>();
            for(int i = 0; i < curLevel; i++){
                TreeNode tmp = queue.poll();
                lvl.add(tmp.val);
                if(tmp.left != null){
                    queue.add(tmp.left);
                    nextLevel ++;
                }
                if(tmp.right != null){
                    queue.add(tmp.right);
                    nextLevel ++;
                }
            }
            result.add(lvl);
            
        }
        return result;
    }
}