mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-06 17:44:10 +08:00
57 lines
2.6 KiB
Markdown
Executable File
57 lines
2.6 KiB
Markdown
Executable File
# [1028. Recover a Tree From Preorder Traversal](https://leetcode.com/problems/recover-a-tree-from-preorder-traversal/)
|
||
|
||
|
||
## 题目
|
||
|
||
We run a preorder depth first search 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.
|
||
|
||
Given the output `S` of this traversal, recover the tree and return its `root`.
|
||
|
||
**Example 1:**
|
||
|
||
|
||
|
||
Input: "1-2--3--4-5--6--7"
|
||
Output: [1,2,5,3,4,6,7]
|
||
|
||
**Example 2:**
|
||
|
||
|
||
|
||
Input: "1-2--3---4-5--6---7"
|
||
Output: [1,2,5,3,null,6,null,4,null,7]
|
||
|
||
**Example 3:**
|
||
|
||
|
||
|
||
Input: "1-401--349---90--88"
|
||
Output: [1,401,null,349,88,90]
|
||
|
||
**Note:**
|
||
|
||
- The number of nodes in the original tree is between `1` and `1000`.
|
||
- Each node will have a value between `1` and `10^9`.
|
||
|
||
## 题目大意
|
||
|
||
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。给出遍历输出 S，还原树并返回其根节点 root。
|
||
|
||
|
||
提示：
|
||
|
||
- 原始树中的节点数介于 1 和 1000 之间。
|
||
- 每个节点的值介于 1 和 10 ^ 9 之间。
|
||
|
||
|
||
## 解题思路
|
||
|
||
- 给出一个字符串，字符串是一个树的先根遍历的结果，其中破折号的个数代表层数。请根据这个字符串生成对应的树。
|
||
- 这一题解题思路比较明确，用 DFS 就可以解题。边深搜字符串，边根据破折号的个数判断当前节点是否属于本层。如果不属于本层，回溯到之前的根节点，添加叶子节点以后再继续深搜。需要注意的是每次深搜时，扫描字符串的 index 需要一直保留，回溯也需要用到这个 index。
|