[Leetcode]@python 106. Construct Binary Tree from Inorder and Postorder Traversal

题目链接

https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

题目原文

Given inorder and postorder traversal of a tree, construct the binary tree.

题目大意

根据中序遍历和后序遍历构建二叉树

解题思路

假设后序串:AEFDHZMG,后序的最后一个节点一定是根节点,这样我们就知道了根节点是G. 再看中序ADEFGHMZ, 在中序串之中,根结点的前边的所有节点都是左子树中,所以G节点前面的ADEF就是左子树的中序串。再看后序串 AEFDHZMG, 由于左子树的节点是ADEF,我们可以得到左子树的后序串为: AEFD. 有了左子树的后序串AEFD,和中序串ADEF ,我们就可以递归的把左子树给建立起来。 同样,可以建立起右子树。(但是这里要先构建右子树,再构建左子树,至于原因为啥,我还不造)

代码

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def buildTree(self, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        if not inorder:
            return None

        index = inorder.index(postorder.pop())
        root = TreeNode(inorder[index])

        root.right = self.buildTree(inorder[index + 1:len(inorder)], postorder)
        root.left = self.buildTree(inorder[0:index], postorder)
        
        return root  
posted @ 2016-03-10 17:08  slurm  阅读(304)  评论(0编辑  收藏  举报