105. Construct Binary Tree from Preorder and Inorder Traversal

Medium
Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.
Example 1:
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​
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output:
[3,9,20,null,null,15,7]
Example 2:
Input: preorder = [-1], inorder = [-1]
Output:
[-1]
Constraints:
  • 1 <= preorder.length <= 3000
  • inorder.length == preorder.length
  • -3000 <= preorder[i], inorder[i] <= 3000
  • preorder and inorder consist of unique values.
  • Each value of inorder also appears in preorder.
  • preorder is guaranteed to be the preorder traversal of the tree.
  • inorder is guaranteed to be the inorder traversal of the tree.

解題

preorder 陣列中,中間節點會在最前面。
inorder 陣列中,中間節點左邊為左子樹,右邊為右子樹。
所以我們可以每次將 preorder 陣列第一個元素抓出來,在 inorder 搜尋該元素,將元素左半進行遞迴呼叫視為元素的左子樹,右半則是右子樹。
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func buildTree(preorder []int, inorder []int) *TreeNode {
if len(preorder) == 0 {
return nil
}
​
node := &TreeNode{ Val:preorder[0] }
​
if len(preorder) == 1 {
return node
}
​
i := findIndex(inorder, preorder[0])
node.Left = buildTree(preorder[1:i+1], inorder[0: i])
node.Right = buildTree(preorder[i+1:], inorder[i+1:])
​
return node
}
​
func findIndex(arr []int, target int) int {
for i, n := range arr {
if n == target {
return i
}
}
​
return -1
}