二叉树的递归遍历
- 递归三部曲
- 1.确定递归函数的参数和返回值 2.确定终止条件 3.确定单层的递归逻辑
- 前中后序遍历只需要改一下位置即可
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> vec;
traversal(root, vec);
return vec;
}
void traversal(TreeNode* cur, vector<int>& vec) {
if (cur == nullptr) {
return;
}
traversal(cur->left, vec);
vec.push_back(cur->val);
traversal(cur->right, vec);
}
};
二叉树的迭代遍历
- 利用栈的特性
- 栈是先进后出,所以遍历左右节点的时候需要反过来
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
stack<TreeNode*> st;
vector<int> result;
if (root == nullptr) {
return result;
}
st.push(root);
while (!st.empty()) {
TreeNode* cur = st.top();
if (cur != nullptr) {
st.pop();
if (cur->right)
st.push(cur->right);
st.push(cur);
st.push(nullptr);
if (cur->left)
st.push(cur->left);
} else {
st.pop();
cur = st.top();
st.pop();
result.push_back(cur->val);
}
}
return result;
}
};
二叉树的层序遍历
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
queue<TreeNode*> que;
vector<vector<int>> result;
if (root == nullptr) {
return result;
}
que.push(root);
while (!que.empty()) {
int size = que.size();
vector<int> vec;
for (int i = 0; i < size; i++) {
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);
}
result.push_back(vec);
}
return result;
}
};
层序遍历刷题