LeetCode 102. 二叉树的层序遍历 && 剑指 Offer 32

给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）。
 

示例：
二叉树：[3,9,20,null,null,15,7],
    3
   / 
  9  20
    /  
   15   7
返回其层序遍历结果：

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> ans;
    int curr = 0;
    vector<int> v;
    void bfs(TreeNode* root){
        if(root == NULL) return;
        queue<pair<TreeNode*,int>> q;
        q.push({root,0});

        while(!q.empty()){
            TreeNode* p = q.front().first; 
            int level = q.front().second; 
            q.pop();

            if(level > curr){
                curr=level; 
                ans.push_back(v);
                v.clear();
                v.push_back(p->val);
            }else{
                v.push_back(p->val);
            }

            if(p->left!=NULL) q.push({p->left,level+1});
            if(p->right!=NULL) q.push({p->right,level+1});
        }
        ans.push_back(v);

        return;
    }

    vector<vector<int>> levelOrder(TreeNode* root) {
        bfs(root);
        return ans;
    }
};

class Solution {
public:
    vector<vector<int>> ans;
    void dfs(TreeNode* root,int level){
        if(root==NULL) return;
        
        if(ans.size() < level){
            vector<int> v{root->val};
            ans.push_back(v);
        }else{
            ans[level-1].push_back(root->val);
        }
        
        dfs(root->left,level+1);dfs(root->right,level+1);
    }
    
    vector<vector<int>> levelOrder(TreeNode* root) {
        dfs(root,1);
        
        return ans;
    }
};