题目描述
思路分析
递归迭代均可,递归加上level显示当前深度
代码展示
/**
* 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>> ret;
if (root == NULL) {
return ret;
}
_help(ret, root, 1);
return ret;
}
void _help(vector<vector<int>>& ret, TreeNode* root, int level) {
if (root == NULL) {
return;
}
if (ret.size() < level) {
ret.push_back(vector<int>());
}
ret[level-1].push_back(root->val);
_help(ret, root->left, level+1);
_help(ret, root->right, level+1);
}
};
结果分析
时间复杂度O(N)
空间复杂度O(N)