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:

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
}