Leetcode 107 Binary Tree Level Order Traversal II 二叉树+BFS

题意是倒过来层次遍历二叉树

下面我介绍下BFS的基本框架，所有的BFS都是这样写的

struct Nodetype {
      int d;//层数即遍历深度
      KeyType m;//相应的节点值
}
queue<Nodetype> q;
q.push(firstnode);
while(!q.empty()){
        Nodetype now = q.front();
         q.pop();
         ........
         for(遍历所有now点的相邻点next){
               if(!visit[next]) {
                   访问每个没有访问过的点;
                   做相应的操作；
                   next.d = now.d + ;
                   q.push(next);
               }
         }
  }

对于二叉树将val看成层数，这边我偷懒了。

 /**
  * 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>> levelOrderBottom(TreeNode* root) {
 
         vector<vector<int>> v;
         if(!root) return v;
         vector<int> t;
 
         t.push_back(root->val);
         v.push_back(t);
         root->val = ;
 
         queue<TreeNode*> q;
         q.push(root);
 
         while(!q.empty()){
             TreeNode* now = q.front();
             q.pop();
             if(!now) continue;
 
             q.push(now->left);
             q.push(now->right);
             if(now->val < v.size()){
                 if(now->left) v[now->val].push_back(now->left->val);
                 if(now->right) v[now->val].push_back(now->right->val);
             }
             else{
                 vector<int> t;
                 if(now->left) t.push_back(now->left->val);
                 if(now->right) t.push_back(now->right->val);
                 if(t.size() != )v.push_back(t);
             }
             if(now->left) now->left->val = now->val + ;
             if(now->right)now->right->val = now->val + ;
         }
         reverse(v.begin(),v.end());
         return v;
     }
 };