题目描述
给定两个整数数组 preorder 和 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。
示例 1:
输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] 输出: [3,9,20,null,null,15,7]
示例 2:
输入: preorder = [-1], inorder = [-1] 输出: [-1]
思路
剑指offer重建二叉树-CSDN博客
1）递归地按照前序的顺序，制作成根节点
2）在中序中找到根节点的索引，按照其索引将树分为左子树和右子树
3）直到前序和中序序列没有
4）返回根节点
# Definition for a binary tree node.
class TreeNode(object):def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right
class Solution(object):def buildTree(self, preorder, inorder):""":type preorder: List[int]:type inorder: List[int]:rtype: TreeNode"""if not preorder or not inorder:return Noneroot = TreeNode(preorder[0])index = inorder.index(root.val)root.left = self.buildTree(preorder[1:index+1],inorder[:index])root.right = self.buildTree(preorder[index+1:],inorder[index+1:])return rootdef printTreeMidOrder(root, res):if not root:return Noneif root.left:printTreeMidOrder(root.left, res)res.append(root.val)if root.right:printTreeMidOrder(root.right, res)if __name__ == '__main__':s = Solution()pre = [1, 2, 4, 7, 3, 5, 6, 8]tin = [4, 7, 2, 1, 5, 3, 8, 6]root = s.buildTree(pre, tin)# 打印中序遍历结果res = []printTreeMidOrder(root, res)print(res)
