Binary Tree Level Order Traversal II 从下向上按层遍历二叉树

Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order

For example:

Given binary tree ​​{3,9,20,#,#,15,7}​​,

3 / \ 9 20 / \ 15 7

return its bottom-up level order traversal as:

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

confused what ​​"{1,#,2,3}"​​ means? ​​​> read more on how binary tree is serialized on OJ.​​

/**
 * Definition for binary tree
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > vec;
        queue<TreeNode *> cur;
        if(root==NULL)
            return vec;

        cur.push(root);
        
        vector<int> vv; 
        queue<TreeNode *> next;
        
        while(!cur.empty())
        {    
            TreeNode *t = cur.front();
            cur.pop();
            vv.push_back(t->val);

            if(t->left)
                next.push(t->left);
            if(t->right)
                next.push(t->right);
                
            if(cur.empty())
            {
                 vec.insert(vec.begin(),vv);//vv从头插入
                // cur=next;
                //next.clear();
                swap(cur,next);
                 vv.clear();
            }
        }
        
        return vec;
    
    }
};