144. Binary Tree Preorder Traversal

Problem Statement

144. Binary Tree Preorder Traversal

Easy

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]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [1,2,4,5,6,7,3,8,9]

Explanation:

Example 3:

Input: root = []

Output: []

Example 4:

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?

C++

Source file

/**
 * 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> v;
    vector<int> preorderTraversal(TreeNode* root) {
        if(root){
            v.push_back(root->val);
            preorderTraversal(root->left);
            preorderTraversal(root->right);
        }
        return v;
    }
};

Java

Source file

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> pre = new ArrayList<>();
        solve(root, pre);
        return pre;
    }

    void solve(TreeNode node, List<Integer> pre) {
        if (node == null)
            return;
        pre.add(node.val);
        solve(node.left, pre);
        solve(node.right, pre);
    }
}