LeetCode OJ--Binary Tree Level Order Traversal

http://oj.leetcode.com/problems/binary-tree-level-order-traversal/

树的层序遍历，使用队列

由于树不是满的，还要分出每一层来，刚开始给缺少的节点用dummy节点代替，结果超时了。

vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int> > ans;
        if(root == NULL)
            return ans;
        int num = 1,num2 = 1;
        queue<TreeNode *> myQueue;
        myQueue.push(root);
        TreeNode *nodeFront;
        TreeNode *dummy = new TreeNode(-1);

        vector<int> onePiece;
        while(!myQueue.empty())
        {
            nodeFront = myQueue.front();
            myQueue.pop();
            num--;
            if(nodeFront != dummy)
            {
                onePiece.push_back(nodeFront->val);
                if(nodeFront->left)
                    myQueue.push(nodeFront->left);
                else
                    myQueue.push(dummy);
                if(nodeFront->right)
                    myQueue.push(nodeFront->right);
                else
                    myQueue.push(dummy);
            }
            else
            {
                myQueue.push(dummy);
                myQueue.push(dummy);
            }
            
            
            if(num == 0)
            {
                if(onePiece.empty())
                break;
                ans.push_back(onePiece);
                onePiece.clear();
                num2 = num2*2;
                num = num2;
            }
        }
        return ans;
    }

改进的话，对缺失的节点进行计数，则计算出下一层应该有多少个节点来，如下。

    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int> > ans;
        if(root == NULL)
            return ans;
        int num = 1,num2 = 1,nullNum = 0,nullNumAcc = 0;
        queue<TreeNode *> myQueue;
        myQueue.push(root);
        TreeNode *nodeFront;

        vector<int> onePiece;
        while(!myQueue.empty())
        {
            nodeFront = myQueue.front();
            myQueue.pop();
            num--;
             
            onePiece.push_back(nodeFront->val);
            if(nodeFront->left)
                myQueue.push(nodeFront->left);
            else
                nullNum++;
            if(nodeFront->right)
                myQueue.push(nodeFront->right);
            else
                nullNum++;
             
            if(num == 0)
            {
                if(onePiece.empty())
                    break;
                ans.push_back(onePiece);
                onePiece.clear();
                num2 = num2*2;
                nullNumAcc = nullNumAcc*2 + nullNum;
                num = num2 - nullNumAcc; 
                nullNum = 0;
            }
        }
        return ans;
    }