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,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
/**二叉树的bfs和队列个数,可分层处理到每一层的的节点。
* Definition for a binary tree node.
* 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) {
vector<vector<int> > outer;
vector<int> inner;
queue<TreeNode*> q;
if (root)
q.push(root);
else
return outer;
while (!q.empty())
{
int size=q.size();//当前队列的个数,用来将当前层的节点全从队列中移除。
inner.clear();
while (size-->0)
{
TreeNode* t=q.front();
q.pop();
inner.push_back(t->val);
if (t->left)
q.push(t->left);
if (t->right)
q.push(t->right);
}
outer.push_back(inner);
}
return outer;
}
};