题意:
对二叉树按层遍历。每一层一行。
思路:
bfs。每次要打印一个结点时,先把它的子节点加入到队列当中 。
用两个数字计数表示当前层和下一层的剩余结点数和当前结点数。
做过了。还是不会。
代码:
/**
* 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) {
if (root == NULL) return {};
queue<TreeNode*> que;
que.push(root);
int curcnt = 1;
int nxtcnt = 0;
vector<vector<int>> ans;
vector<int> temp;
while(!que.empty()) {
TreeNode* cur = que.front();
que.pop();
temp.push_back(cur->val);
if (cur->left) {
que.push(cur->left);
nxtcnt++;
}
if (cur->right) {
que.push(cur->right);
nxtcnt++;
}
curcnt--;
if (curcnt == 0) {
ans.push_back(temp);
temp.clear();
curcnt = nxtcnt;
nxtcnt = 0;
}
}
return ans;
}
};