难度中等509
给你二叉树的根节点
root
,返回它节点值的 前序 遍历。示例 1:
输入:root = [1,null,2,3] 输出:[1,2,3]示例 2:
输入:root = [] 输出:[]示例 3:
输入:root = [1] 输出:[1]示例 4:
输入:root = [1,2] 输出:[1,2]示例 5:
输入:root = [1,null,2] 输出:[1,2]提示:
- 树中节点数目在范围
[0, 100]
内-100 <= Node.val <= 100
进阶:递归算法很简单,你可以通过迭代算法完成吗?
1、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 preTraverse(TreeNode* root, vector<int> & res){
if(root == nullptr){
return ;
}
res.push_back(root->val);
preTraverse(root->left, res);
preTraverse(root->right, res);
}
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
preTraverse(root, res);
return res;
}
};
2 、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) {
vector<int> res;
if(root == nullptr){
return res;
}
stack<TreeNode*> stk;
while(root != nullptr || !stk.empty()){
while(root != nullptr){
res.push_back(root->val);
stk.push(root);
root = root->left;
}
root = stk.top();
stk.pop();
root = root->right;
}
return res;
}
};