I
菜鸡思路：
数组的层次遍历，放到数组里即可。
菜鸡代码：
/**
* 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<int> levelOrder(TreeNode* root) {
vector<int> result;
queue<TreeNode*> que;
if(root != NULL) {
que.push(root);
}
while(!que.empty()) {
int size = que.size();
for(int i = 0; i < size; i++) {
TreeNode* node = que.front();
que.pop();
result.push_back(node->val);
if(node->left != NULL) {
que.push(node->left);
}
if(node->right != NULL) {
que.push(node->right);
}
}
}
return result;
}
};
II
菜鸡思路：
虽然还是层次遍历，但是要求分层返回，有些棘手，那么可以这样，每次判断一下一层有几个，然后针对性返回该层。
菜鸡代码：
/**
* 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) {
vector<vector<int>> result;
queue<TreeNode *> que;
if (root) {
que.push(root);
}
while (que.size()) {
vector<int> row;
int size = que.size();
//记录下这一层有几个
for (int i = 0; i < size; ++i) {
//针对每层的节点数分层操作
TreeNode *node = que.front();
que.pop();
row.push_back(node->val);
if (node->left) {
que.push(node->left);
}
if (node->right) {
que.push(node->right);
}
}
result.push_back(row);
}
return result;
}
};
III
菜鸡思路：
虽然还是层次遍历、分层返回，但是要求的是要针对性的对一些层倒序，直接在上一题代码改造，判断一下是第几层，偶数层则数组倒序。只比上一题多了一个if条件块。
菜鸡代码：
/**
* 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) {
vector<vector<int>> result;
queue<TreeNode *> que;
if (root) {
que.push(root);
}
while (que.size()) {
vector<int> row;
int size = que.size();
for (int i = 0; i < size; ++i) {
TreeNode *node = que.front();
que.pop();
row.push_back(node->val);
if (node->left) {
que.push(node->left);
}
if (node->right) {
que.push(node->right);
}
}
if (result.size() % 2) {
reverse(row.begin(), row.end());
}
result.push_back(row);
}
return result;
}
};