请看题
Example
思路
这道题太简单了，直接上代码了
Code
/**
* 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> result;
void InordenAux(TreeNode *root)
{
if(!root){return;}
InordenAux(root->left);
result.push_back(root->val);
InordenAux(root->right);
}
vector<int> inorderTraversal(TreeNode* root) {
if(!root)
{
return result;
}
InordenAux(root);
return result;
}
};
代码解析
这一道题非常简单，二叉树无序遍历无非就是递归，只需要一直递归然后保存就好了，太简单了这道题。没啥可说的
结束
Preorden And Postorden
vector<int> result;
void preorderTraversalAux(TreeNode *root)
{
if(!root){return;}
result.push_back(root->val);
preorderTraversalAux(root->left);
preorderTraversalAux(root->right);
}
vector<int> preorderTraversal(TreeNode* root) {
if(!root)
{
return result;
}
preorderTraversalAux(root);
return result;
}
class Solution {
public:
vector<int> result;
void postorderTraversalAux(TreeNode *root)
{
if(!root){return;}
postorderTraversalAux(root->left);
postorderTraversalAux(root->right);
result.push_back(root->val);
}
vector<int> postorderTraversal(TreeNode* root) {
if(!root)
{
return result;
}
postorderTraversalAux(root);
return result;
}
};