• 假如有前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}
  • 从前序pre知根节点是第一个元素1，从中序知元素1前面的[4,7,2]都是左子树，[5,3,8,6]是右子树
  • 递归可建得二叉树 
class Solution:
    # 返回构造的TreeNode根节点
    def reConstructBinaryTree(self, pre, tin):
        # write code here
        if not pre or not tin:
            return None
        root = TreeNode(pre[0])
        k = tin.index(pre[0])
        root.left = self.reConstructBinaryTree(pre[1:k+1],tin[0:k])
        root.right = self.reConstructBinaryTree(pre[k+1:],tin[k+1:])
        return root