重建二叉树—递归

Posted 2020-12-18 gugu-da

二叉树重建

• 问题：输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重 复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

• 解决：

  #递归一
  # class TreeNode:
  #     def __init__(self, x):
  #         self.val = x
  #         self.left = None
  #         self.right = None
  class Solution:
      def reConstructBinaryTreeCore(self,pre,pre_start,pre_end,tin,tin_start,tin_end):
          if pre_start > pre_end or tin_start > tin_end:
              return None
          root = TreeNode(pre[pre_start])
          for i in range(tin_start,tin_end+1):
              if pre[pre_start] == tin[i]:
                  root.left = self.reConstructBinaryTreeCore(pre,pre_start+1,i-tin_start+pre_start,tin,tin_start,i-1)
                  root.right = self.reConstructBinaryTreeCore(pre,i-tin_start+pre_start+1,pre_end,tin,i+1,tin_end)
                  break
          return root
      # 返回构造的TreeNode根节点
      def reConstructBinaryTree(self, pre, tin):
          # write code here
          if len(pre)==0 or len(tin)==0:
              return None
          return self.reConstructBinaryTreeCore(pre,0,len(pre)-1,tin,0,len(tin)-1)
  #递归二
  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
          root = TreeNode(pre[0])
          i = tin.index(pre[0])
          root.left =  self.reConstructBinaryTree(pre[1:i+1], tin[:i])
          root.right = self.reConstructBinaryTree(pre[i+1:], tin[i+1:])
          return root