Binary Tree Preorder Traversal
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?
Solutions
🧠 Cpp
/**
* 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:
vector<int> preorderTraversal(TreeNode* root)
{
vector<int> res;
if(root)
res.emplace_back(root->val);
else
return res;
if(root->left)
{
vector<int> &&lres = preorderTraversal(root->left);
res.insert(res.end(), lres.begin(), lres.end());
}
if(root->right)
{
vector<int> &&lres = preorderTraversal(root->right);
res.insert(res.end(), lres.begin(), lres.end());
}
return res;
}
};
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