"""
按照中根遍历的顺序构建BST
:param head: ListNode
:param begin: int
:param end: int
:return: TreeNode
"""
if begin > end:
return None
mid = (begin+end)//2
leftTree = self.buildBST(begin,mid-1)
root = TreeNode(self.node.val)
root.left = leftTree
self.node = self.node.next
root.right = self.buildBST(mid+1,end)
return root
if __name__ == '__main__':
so = Solution()
head = ListNode.getOneListNode([1,2,3,4,5])
root = so.sortedListToBST(head)
TreeNode.inOrderRec(root)
return None
return self.buildTreeFromInPost(inorder,0,len(inorder)-1,postorder,0,len(postorder)-1)
def buildTreeFromInPost(self,inorder,iB,iE,postorder,pB,pE):
"""
根据中根和后根遍历构建二叉树。后跟遍历的最后一个是根。
:param inorder: List[int]
:param iB:int
:param iE:int
:param postorder: List[int]
:param pB: int
:param pE:int
:return:TreeNode
"""
if iB > iE or pB > pE:return None
if iB == iE:
return TreeNode(inorder[iE])
root = TreeNode(postorder[pE])
mid = inorder.index(postorder[pE])
leftLen = mid - iB
root.left = self.buildTreeFromInPost(inorder,iB,mid-1,postorder,pB,pB+leftLen-1)
root.right = self.buildTreeFromInPost(inorder,mid+1,iE,postorder,pB+leftLen,pE-1)
return root
if __name__ == '__main__':
so = Solution()
root = so.buildTree([0,1,2,3,4,5,6,],[0,2,1,4,6,5,3])
TreeNode.inOrderRec(root)
"""
根据先根和中根还原二叉树。
:type preorder: List[int]
:type inorder: List[int]
:rtype: TreeNode
"""
if not preorder:
return None
return self.buildTreeFromPreIn(preorder,0,len(preorder)-1,inorder,0,len(inorder)-1)
def buildTreeFromPreIn(self,preorder,pB,pE,inorder,iB,iE):
if pB > pE or iB > iE:
return None
if pB == pE:
return TreeNode(preorder[pE])
root = TreeNode(preorder[pB])
mid = 0
for i in range(iB,iE+1):
if inorder[i] == preorder[pB]:
mid = i
break
leftSize = mid-iB
root.left = self.buildTreeFromPreIn(preorder,pB+1,pB+leftSize,inorder,iB,mid-1)
root.right = self.buildTreeFromPreIn(preorder,pB+leftSize+1,pE,inorder,mid+1,iE)
return root
if __name__ == '__main__':
so = Solution()
TreeNode.inOrderRec(so.buildTree([1,2,4,3],[4,2,1,3]))
