/**
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
class Solution {
public:
vector<int> result;
/**
*
* @param root TreeNode类
* @return int整型vector
*/
vector<int> preorderTraversal(TreeNode* root) {
// write code here
if(root == NULL)
return result;
stack<TreeNode*> s;
s.push(root);
while(!s.empty())
{
TreeNode* node = s.top();
s.pop();
result.push_back(node->val);
if(node->right)
s.push(node->right);
if(node->left)
s.push(node->left);
}
return result;
}
};
这个是比较经典的递归问题,可以看下代码,如果这个代码看不懂,可以私聊我,我拉你进来打卡学习
void preorder(struct TreeNode* root, int *ret, int *returnSize) {
if(root) {
ret[ (*returnSize) ++ ] = root->val; // (1) 先访问根结点;
preorder(root->left, ret, returnSize); // (2) 再前序遍历左子树;
preorder(root->right, ret, returnSize); // (3) 最后前序遍历右子树;
}
}
int* preorderTraversal(struct TreeNode* root, int* returnSize){
int *ret = (int *) malloc( sizeof(int) * 105 ); // (4) 数组内存申请的时候比最大个数多一点无妨;
*returnSize = 0; // (5) 数组长度置0;
preorder(root, ret, returnSize); // (6) 开始前序遍历,并且记录结果到```ret```中;
return ret;
}