102. 二叉树的层序遍历
题目描述
给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。
解题思路
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> res ;
if (root == NULL) return res ;
deque<TreeNode*> v;
v.push_back(root);
while (!v.empty()) {
int size = v.size();
vector<int> level_res ;
for (int i = 0; i < size; i ++) {
TreeNode * ptr = v.front();
v.pop_front();
level_res.push_back(ptr->val);
if (ptr->left) v.push_back(ptr->left);
if (ptr->right) v.push_back(ptr->right);
}
res.push_back(level_res) ;
}
return res;
}
};
学习感想
107. 二叉树的层序遍历 II
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> res ;
if (root == NULL) return res ;
deque<TreeNode*> v;
v.push_back(root);
while (!v.empty()) {
int size = v.size();
vector<int> level_res ;
for (int i = 0; i < size; i ++) {
TreeNode * ptr = v.front();
v.pop_front();
level_res.push_back(ptr->val);
if (ptr->left) v.push_back(ptr->left);
if (ptr->right) v.push_back(ptr->right);
}
res.push_back(level_res) ;
}
reverse(res.begin(),res.end());
return res;
}
};
199. 二叉树的右视图
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<int> res ;
if (root == NULL) return res ;
deque<TreeNode*> v;
v.push_back(root);
while (!v.empty()) {
int size = v.size();
vector<int> level_res ;
for (int i = 0; i < size; i ++) {
TreeNode * ptr = v.front();
v.pop_front();
level_res.push_back(ptr->val);
if (ptr->left) v.push_back(ptr->left);
if (ptr->right) v.push_back(ptr->right);
}
res.push_back(level_res.back());
}
return res;
}
};