Given the root of a binary tree, return the inorder traversal of its nodes’ values.
Example 1:
Input: root = [1,null,2,3]
Output: [1,3,2]
Explanation:
Example 2:
Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]
Output: [4,2,6,5,7,1,3,9,8]
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?
Approach - Recursion
class Solution {
List<Integer> ans = new ArrayList<>();
public List<Integer> inorderTraversal(TreeNode root) {
dfs(root);
return ans;
}
void dfs (TreeNode root) {
if (root == null) return;
dfs(root.left);
ans.add(root.val);
dfs(root.right);
}
}