[LeetCode 94] Binary Tree Inorder Traversal

문제 바로가기

Description

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Given the root of a binary tree, return the inorder traversal of its nodes’ values.

Example 1

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• Input: root = [1,null,2,3]
• Output: [1,3,2]
• Explanation:
  • 

Example 2

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• 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

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• Input: root = []
• Output: []

Example 4

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• Input: root = [1]
• Output: [1]

Constraints

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• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

Code

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내 제출

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/**
 * 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 {
    private List<Integer> ans = new ArrayList<>();

    public List<Integer> inorderTraversal(TreeNode root) {
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        ans.add(root.val);
        dfs(root.right);
    }
}

Runtime	Memory
0 ms	41.7 MB

다른 풀이

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class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        inorder(root, res);
        return res;
    }

    private void inorder(TreeNode root, List<Integer> res) {
        if (root == null) return;          // base case

        inorder(root.left, res);           // go LEFT
        res.add(root.val);                 // visit ROOT
        inorder(root.right, res);          // go RIGHT
    }
}

Reference

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• https://github.com/doocs/leetcode/blob/main/solution/0000-0099/0094.Binary%20Tree%20Inorder%20Traversal/Solution.java