102. Binary Tree Level Order Traversal. (Medium)

Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).
 
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
Example 2:
Input: root = [1]
Output: [[1]]
Example 3:
Input: root = []
Output: []
 
Constraints:
The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

A：

 ---------------简单的dfs， 每当看见一个节点，就加到结果上-----------------

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        helper(res, root, 0);
        return res;
    }
private:
    void helper(vector<vector<int>>& res, TreeNode* root, int depth) {
        if (!root)
            return;
        if (depth == res.size()) {
            vector<int> a;
            res.push_back(a);
        }
        res[depth].push_back(root->val);
        helper(res, root->left, depth + 1);
        helper(res, root->right, depth + 1);
    }
};

----------------------------第二遍，-------每次取下来一层，   放到一个Node 的list里，  同时保留Integer的一层---------------------

public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> all = new LinkedList();
        if(root == null)
            return all;
        List<TreeNode> lastLevel = new LinkedList();
        List<Integer> l = new LinkedList();
        lastLevel.add(root);
        l.add(root.val);
        all.add(l);
        while(lastLevel.isEmpty() ==false){
            l = new LinkedList();
            List<TreeNode> newLevel = new LinkedList();
            for(TreeNode node:lastLevel){
                if( node.left !=null ){
                    newLevel.add(node.left);
                    l.add(node.left.val);
                }
                if( node.right !=null){
                    newLevel.add(node.right);
                    l.add(node.right.val);
                }
            }
            lastLevel = newLevel;
            if(l.isEmpty()==false)
                all.add(l);
        }
        return all;
    }
}

Mistakes:
 1: 考虑到 stack的特性， 我们在加入的时候，要先push right child ,then left
 2  艹～～～～～～～～～  BFS 是用Queue， 而不是stack， TNND
 3：  每次要在while里重新new一个List， 单纯的clear（）的话， ----由于其指针特性，会把all里的结果都clear掉
Learned:
1:   经常出现错误说“cannot create a generic array of .......”

Because of how generics in Java work, you cannot directly create an array of a generic type (such as Map<String, Object>[]). Instead, you create an array of the raw type (Map[]) and cast it to Map<String, Object>[]. This will cause an unavoidable (but suppressible) compiler warning. This should work for what you need: Map<String, Object>[] myArray = (Map<String, Object>[]) new Map[10]; You may want to annotate the method this occurs in with @SupressWarnings("unchecked"), to prevent the warning from being shown.

还有一个办法，就是  在new 的时候，不加generic type， 例如： 直接也不进行上面的cast
        Stack<TreeNode>[] twoStack = new Stack[2];

3： Queue is an Interface not a class. ，  so,we cannot use them as class.
Instead, we need use LinkedList or PriorityQueue and etc to simulate it.