![\"\"](\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/04/12/recover-a-tree-from-preorder-traversal.png\")
We run a preorder depth-first search (DFS) on the root of a binary tree.
At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node. If the depth of a node is D, the depth of its immediate child is D + 1. The depth of the root node is 0.
If a node has only one child, that child is guaranteed to be the left child.
\n\nGiven the output traversal of this traversal, recover the tree and return its root.
\n
Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2024/09/10/recover_tree_ex1.png\")
\n\nInput: traversal = "1-2--3--4-5--6--7"\nOutput: [1,2,5,3,4,6,7]\n\n\n
Example 2:
\n![\"\"](\"https://assets.leetcode.com/uploads/2024/09/10/recover_tree_ex2.png\")
\n\nInput: traversal = "1-2--3---4-5--6---7"\nOutput: [1,2,5,3,null,6,null,4,null,7]\n\n\n
Example 3:
\n![\"\"](\"https://assets.leetcode.com/uploads/2024/09/10/recover_tree_ex3.png\")
\n\nInput: traversal = "1-401--349---90--88"\nOutput: [1,401,null,349,88,90]\n\n\n
\n
Constraints:
\n\n[1, 1000].1 <= Node.val <= 109我们从二叉树的根节点 root 开始进行深度优先搜索。
在遍历中的每个节点处，我们输出 D 条短划线（其中 D 是该节点的深度），然后输出该节点的值。（如果节点的深度为 D，则其直接子节点的深度为 D + 1。根节点的深度为 0）。
如果节点只有一个子节点，那么保证该子节点为左子节点。
\n\n给出遍历输出 S，还原树并返回其根节点 root。
\n\n
示例 1：
\n\n
输入："1-2--3--4-5--6--7"\n输出：[1,2,5,3,4,6,7]\n\n\n
示例 2：
\n\n![\"\"](\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/04/12/screen-shot-2019-04-10-at-114101-pm.png\")
输入："1-2--3---4-5--6---7"\n输出：[1,2,5,3,null,6,null,4,null,7]\n\n\n
示例 3：
\n\n![\"\"](\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/04/12/screen-shot-2019-04-10-at-114955-pm.png\")
输入："1-401--349---90--88"\n输出：[1,401,null,349,88,90]\n\n\n
\n\n
提示：
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"style": "LEETCODE",
"exampleTestcases": "\"1-2--3--4-5--6--7\"\n\"1-2--3---4-5--6---7\"\n\"1-401--349---90--88\"",
"__typename": "QuestionNode"
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