根据一棵树的中序遍历与后序遍历构造二叉树。

中序遍历 inorder = [9,3,15,20,7]后序遍历 postorder = [9,15,7,20,3]

3   / \  9  20    /  \   15   7

Python 实现

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def buildTree(self, inorder, postorder):        """        :type inorder: List[int]        :type postorder: List[int]        :rtype: TreeNode        """        if not postorder or not inorder:            return None;        node = TreeNode(postorder[-1])        mid = inorder.index(postorder[-1])        node.left = self.buildTree(inorder[:mid],postorder[:mid])        node.right = self.buildTree(inorder[mid+1:],postorder[mid:-1])        return node