94. Binary Tree Inorder Traversal

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DFS

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class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        // left -> root -> right

        List<Integer> list = new LinkedList<>();
        recur(list, root);
        return list;
    }

    private void recur(List<Integer> list, TreeNode node) {
        if (node == null) {
            return;
        }

        recur(list, node.left);
        list.add(node.val);
        recur(list, node.right);
    }
}

Iterate

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class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        // left -> root -> right

        List<Integer> resultList = new ArrayList<>();

        Stack<TreeNode> stack = new Stack<>();
        while (root != null || !stack.isEmpty()) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }

            // now: root is null
            TreeNode node = stack.pop(); // 回溯
            resultList.add(node.val);
            if (node.right != null) {
                root = node.right; // 注意此处是赋值给 root 而不是 stack.push(node.right)，因为需要将 node.right 作为一颗子树进行递归
            }
        }

        return resultList;
    }
}

References

94. Binary Tree Inorder Traversal