87. Binary Tree Inorder Traversal

Inorder visits the left subtree, then the node, then the right subtree — for a BST this yields sorted order.

Left, Node, Right

dfs(node):
    if node is None: return
    dfs(node.left)
    visit node.val
    dfs(node.right)

Worked example

[1,null,2,3] (1 with right child 2, whose left child is 3) → [1,3,2].

Complexity

• Time: O(n).
• Space: O(h) recursion.

import java.util.ArrayList;
import java.util.List;

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

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