面试算法-96-二叉树展开为链表-解

class Solution {
    public void flatten(TreeNode root) {
        dfs(root);
    }

    TreeNode pre = null;

    public void dfs(TreeNode root) {
        if (root == null) {
            return;
        }

        dfs(root.right);
        dfs(root.left);

        root.right = pre;
        root.left = null;
        pre = root;
    }
}