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思路和依据前序遍历和中序遍历重建树的思路一样，复杂度也一致，代码如下：
class Solution(object): def buildTree(self, inorder, postorder): """ :type inorder: List[int] :type postorder: List[int] :rtype: TreeNode """ if not inorder: return None length = len(inorder) map = {} for i in xrange(length): map[inorder[i]] = i return self.helper(postorder,0,length-1,0,length-1,map) def helper(self,postorder,Istart,Iend,Pstart,Pend,map): if Istart > Iend: return None node = TreeNode(postorder[Pend]) if Istart == Iend: return node index = map[node.val] node.left = self.helper(postorder,Istart,index-1, Pstart,Pstart+index-Istart-1,map) node.right = self.helper(postorder,index+1,Iend,Pstart+index-Istart,Pend-1,map) return node
Construct Binary Tree from Inorder and Postorder Traversal
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原文地址：http://www.cnblogs.com/sherylwang/p/5405724.html
