144. Binary Tree Preorder Traversal

Description

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]

 

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?

Solution

binary-tree-preorder-traversal.py

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []

        def go(node):
            if not node: return

            res.append(node.val)
            go(node.left)
            go(node.right)

        go(root)
        return res

binary-tree-preorder-traversal.cpp

/**
 * 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> res;

    void helperR(TreeNode* root){
        if (root == nullptr) return;

        res.push_back(root->val);
        helperR(root->left);
        helperR(root->right);
    }

    vector<int> preorderTraversal(TreeNode* root) {
        helperR(root);

        return res;
    }
};