LeetCode | 144. Binary Tree Preorder Traversal
2020-10-02 18:00:03
題目：
Given the root of a binary tree, return the preorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3] Output: [1,2,3]
Example 2:
Input: root = [] Output: []
Example 3:
Input: root = [1] Output: [1]
Example 4:
Input: root = [1,2] Output: [1,2]
Example 5:
Input: root = [1,null,2] Output: [1,2]
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Follow up:
Recursive solution is trivial, could you do it iteratively?
程式碼：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void backPreorder(TreeNode* root, vector<int>& res) {
if(root == NULL)
return;
res.push_back(root->val);
backPreorder(root->left, res);
backPreorder(root->right, res);
return;
}
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
backPreorder(root, res);
return res;
}
};