立志重刷代码随想录60天冲冲冲!!——第十三天
144. 二叉树的前序遍历
前序:中左右
递归
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
// 确定传入参数、返回值
void traversal (TreeNode* curr, vector<int>& vec) {
// 确定终止条件
if (curr == NULL) return;
// 确定递归逻辑
vec.push_back(curr->val);
traversal(curr->left, vec);
traversal(curr->right, vec);
}
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
traversal(root, res);
return res;
}
};
(迭代)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
// 栈存放结点
stack<TreeNode*> st;
vector<int> res;
if (root == nullptr) return res;
st.push(root);
while (!st.empty()) {
TreeNode* node = st.top();
st.pop();
res.push_back(node->val);
// 前序:中左右。但是因为是栈,所以先放右边,后放左边
if (node->right) st.push(node->right);
if (node->left) st.push(node->left);
}
return res;
}
};
145. 二叉树的后序遍历
前序:左右中
递归
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void Traversal (TreeNode* curr, vector<int>& vec) {
if (curr == nullptr) return;
Traversal(curr->left, vec);
Traversal(curr->right, vec);
vec.push_back(curr->val);
}
vector<int> postorderTraversal(TreeNode* root) {
vector<int> res;
Traversal(root, res);
return res;
}
};
迭代
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
stack<TreeNode*> st;
vector<int> res;
if (root == nullptr) return res;
st.push(root);
while (!st.empty()) {
TreeNode* node = st.top();
st.pop();
res.push_back(node->val);
// 后序:左右中。先左后右,
if (node->left) st.push(node->left);
if (node->right) st.push(node->right);
}
// 再翻转!
reverse(res.begin(), res.end());
return res;
}
};
94. 二叉树的中序遍历
递归
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void Traversal (TreeNode* curr, vector<int>& vec) {
if (curr == nullptr) return;
Traversal(curr->left, vec);
vec.push_back(curr->val);
Traversal(curr->right, vec);
}
vector<int> inorderTraversal(TreeNode* root) {
vector<int> res;
Traversal(root, res);
return res;
}
};
102. 二叉树的层序遍历
重点记!!!!!!
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
queue<TreeNode*> que;
vector<vector<int>> results;
if (root != nullptr) que.push(root);
while (!que.empty()) {
int size = que.size();
vector<int> vec;
while (size--) {
TreeNode* node = que.front();
que.pop();
vec.push_back(node->val);
if (node->left) que.push(node->left);
if (node->right) que.push(node->right);
}
results.push_back(vec);
}
return results;
}
};
107. 二叉树的层序遍历 II
在最后反转数组就行
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
queue<TreeNode*> que;
vector<vector<int>> results;
if (root != nullptr) que.push(root);
while (!que.empty()) {
int size = que.size();
vector<int> vec;
while (size--) {
TreeNode* node = que.front();
que.pop();
vec.push_back(node->val);
if (node->left) que.push(node->left);
if (node->right) que.push(node->right);
}
results.push_back(vec);
}
reverse(results.begin(), results.end());
return results;
}
};
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