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Commitcc1f257

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Create 105. 从前序与中序遍历序列构造二叉树.md
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####105. 从前序与中序遍历序列构造二叉树
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难度:中等
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---
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给定两个整数数组 `preorder``inorder` ,其中 `preorder` 是二叉树的**先序遍历**`inorder` 是同一棵树的**中序遍历** ,请构造二叉树并返回其根节点。
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**示例 1:**
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![](https://assets.leetcode.com/uploads/2021/02/19/tree.jpg)
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```
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输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
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输出: [3,9,20,null,null,15,7]
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```
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**示例 2:**
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```
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输入: preorder = [-1], inorder = [-1]
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输出: [-1]
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```
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**提示:**
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*`1 <= preorder.length <= 3000`
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*`inorder.length == preorder.length`
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*`-3000 <= preorder[i], inorder[i] <= 3000`
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*`preorder` 和 `inorder` 均**无重复** 元素
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*`inorder` 均出现在 `preorder`
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*`preorder` **保证** 为二叉树的前序遍历序列
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*`inorder` **保证** 为二叉树的中序遍历序列
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---
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深度搜索优先:
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每次递归对前序序列和中序序列做切片。具体切到什么位置,拿示例推理一遍。
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```Go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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funcbuildTree(preorder []int,inorder []int) *TreeNode {
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fori:=range inorder {
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if inorder[i] == preorder[0] {
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return &TreeNode{
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Val: inorder[i],
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Left:buildTree(preorder[1:i+1], inorder[:i]),
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Right:buildTree(preorder[i+1:], inorder[i+1:]),
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}
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}
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}
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returnnil
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}
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```

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