leetcode-144-easy

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]
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?

思路一：先序遍历，顺序是 中->左->右，树的遍历用递归是最好实现的

public List<Integer> preorderTraversal(TreeNode root) {
    return preorderTraversal(root, new ArrayList<>());
}

public List<Integer> preorderTraversal(TreeNode root, List<Integer> list) {
    if (root == null) return list;

    list.add(root.val);
    preorderTraversal(root.left, list);
    preorderTraversal(root.right, list);

    return list;
}