Binary Tree Level Order Traversal

 

//Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]


struct QueueNode
{
TreeNode *pTr;
int level;
QueueNode(TreeNode *p,int l):pTr(p),level(l){}
};

class Solution {
public:
vector<vector<int> > levelOrder(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector<int>> result;
queue<QueueNode> que;
int levelcount = -1;

if(root==NULL)
return result;

que.push(QueueNode(root,0));
while(!que.empty())
{
QueueNode queNode = que.front();
que.pop();
if(queNode.level!=levelcount)
{
vector<int> set;
result.push_back(set);
levelcount = queNode.level;
}
result[levelcount].push_back(queNode.pTr->val);
if(queNode.pTr->left)
que.push(QueueNode(queNode.pTr->left,queNode.level+1));
if(queNode.pTr->right)
que.push(QueueNode(queNode.pTr->right,queNode.level+1));
}

return result;
}
};

posted @ 2013-05-25 21:17  summer_zhou  阅读(113)  评论(0)    收藏  举报