1028. Recover a Tree From Preorder Traversal

# 1028. 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 #

• 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。

## 代码 #

``````
package leetcode

import (
"strconv"
)

/**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func recoverFromPreorder(S string) *TreeNode {
if len(S) == 0 {
return &TreeNode{}
}
root, index, level := &TreeNode{}, 0, 0
cur := root
dfsBuildPreorderTree(S, &index, &level, cur)
return root.Right
}

func dfsBuildPreorderTree(S string, index, level *int, cur *TreeNode) (newIndex *int) {
if *index == len(S) {
return index
}
if *index == 0 && *level == 0 {
i := 0
for i = *index; i < len(S); i++ {
if !isDigital(S[i]) {
break
}
}
num, _ := strconv.Atoi(S[*index:i])
tmp := &TreeNode{Val: num, Left: nil, Right: nil}
cur.Right = tmp
nLevel := *level + 1
index = dfsBuildPreorderTree(S, &i, &nLevel, tmp)
index = dfsBuildPreorderTree(S, index, &nLevel, tmp)
}
i := 0
for i = *index; i < len(S); i++ {
if isDigital(S[i]) {
break
}
}
if *level == i-*index {
j := 0
for j = i; j < len(S); j++ {
if !isDigital(S[j]) {
break
}
}
num, _ := strconv.Atoi(S[i:j])
tmp := &TreeNode{Val: num, Left: nil, Right: nil}
if cur.Left == nil {
cur.Left = tmp
nLevel := *level + 1
index = dfsBuildPreorderTree(S, &j, &nLevel, tmp)
index = dfsBuildPreorderTree(S, index, level, cur)
} else if cur.Right == nil {
cur.Right = tmp
nLevel := *level + 1
index = dfsBuildPreorderTree(S, &j, &nLevel, tmp)
index = dfsBuildPreorderTree(S, index, level, cur)
}
}
return index
}

``````

Apr 8, 2023