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;
}