二叉樹重建

• 問題：輸入某二叉樹的前序遍歷和中序遍歷的結果，請重建出該二叉樹。假設輸入的前序遍歷和中序遍歷的結果中都不含重 復的數字。例如輸入前序遍歷序列{1,2,4,7,3,5,6,8}和中序遍歷序列{4,7,2,1,5,3,8,6}，則重建二叉樹並返回。

• 解決：

  #遞迴一
  # 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
      # 返回構造的TreeNode根節點
      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)
  #遞迴二
  class Solution:
      # 返回構造的TreeNode根節點
      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