leetcode-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]
]

confused what 

"{1,#,2,3}"

 means? >
read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as 

"{1,2,3,#,#,4,#,#,5}"

.

二叉树层次遍历，可以用队列解决。

/**
* 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> > levelOrder(TreeNode *root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<vector<int> > ret;
vector<int> sameLevel;

if (root == NULL)
return ret;

queue<NodeInfo> q;

NodeInfo nf(root, 0);

q.push(nf);

int depth = 0;

while (!q.empty()) {
nf = q.front();
sameLevel.clear();

while (nf.depth == depth) {       // 遍历同一层次的节点
sameLevel.push_back((nf.nodePtr)->val);  //把节点值加入vector中

if ((nf.nodePtr)->left != NULL)
q.push(NodeInfo((nf.nodePtr)->left, depth+1));

if ((nf.nodePtr)->right != NULL)
q.push(NodeInfo((nf.nodePtr)->right, depth+1));

q.pop();

if (q.empty())
break;

nf = q.front();
}

ret.push_back(sameLevel);   //把同一层次的数组加入ret中
++depth;
}

//reverse(ret.begin(), ret.end());

return ret;
}
private:
typedef struct NodeInfo {
TreeNode *nodePtr;
int depth;
NodeInfo(TreeNode *ptr, int dep):nodePtr(ptr), depth(dep)
{
}
}NodeInfo;
};