102. 二叉树的层次遍历

给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。

例如:
给定二叉树: [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回其层次遍历结果：

[
  [3],
  [9,20],
  [15,7]
]

dfs解法：

class Solution {
public:
    void dfs(TreeNode *root,int  level,vector<vector<int>> &res)
    {
        if(!root)  return ;
        if(level+1>res.size()) res.push_back({});
        res[level].push_back(root->val);
        dfs(root->left,level+1,res);
        dfs(root->right,level+1,res);
    }
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        dfs(root,0,res);
        return res;
    }
};

bfs解法：

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        TreeNode* right=root,*nextright=NULL;
        queue<TreeNode*> Q;
        if(root) Q.push(root);
        vector<vector<int>> vv;
        vv.push_back(vector<int>());
        while(!Q.empty()){
            TreeNode* cur=Q.front();Q.pop();

            vv.back().push_back(cur->val);

            if(cur->left){Q.push(cur->left);nextright=cur->left;}
            if(cur->right){Q.push(cur->right);nextright=cur->right;}

            if(cur == right){
                right=nextright;
                vv.push_back(vector<int>());
            }
        }

        while(!vv.empty() && vv.back().empty()) vv.pop_back();
        return vv;

    }
};