[leetcode]Construct Binary Tree from Inorder and Postorder Traversal @ Python

原题地址:http://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

题意:根据二叉树的中序遍历和后序遍历恢复二叉树。

解题思路:看到树首先想到要用递归来解题。以这道题为例:如果一颗二叉树为{1,2,3,4,5,6,7},则中序遍历为{4,2,5,1,6,3,7},后序遍历为{4,5,2,6,7,3,1},我们可以反推回去。由于后序遍历的最后一个节点就是树的根。也就是root=1,然后我们在中序遍历中搜索1,可以看到中序遍历的第四个数是1,也就是root。根据中序遍历的定义,1左边的数{4,2,5}就是左子树的中序遍历,1右边的数{6,3,7}就是右子树的中序遍历。而对于后序遍历来讲,一定是先后序遍历完左子树,再后序遍历完右子树,最后遍历根。于是可以推出:{4,5,2}就是左子树的后序遍历,{6,3,7}就是右子树的后序遍历。而我们已经知道{4,2,5}就是左子树的中序遍历,{6,3,7}就是右子树的中序遍历。再进行递归就可以解决问题了。

代码:

[leetcode]Construct Binary Tree from Inorder and Postorder 
Traversal @ Python
# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param inorder, a list of integers
    # @param postorder, a list of integers
    # @return a tree node
    def buildTree(self, inorder, postorder):
        if len(inorder) == 0:
            return None
        if len(inorder) == 1:
            return TreeNode(inorder[0])
        root = TreeNode(postorder[len(postorder) - 1])
        index = inorder.index(postorder[len(postorder) - 1])
        root.left = self.buildTree(inorder[ 0 : index ], postorder[ 0 : index ])
        root.right = self.buildTree(inorder[ index + 1 : len(inorder) ], postorder[ index : len(postorder) - 1 ])
        return root
[leetcode]Construct Binary Tree from Inorder and Postorder 
Traversal @ Python

 

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[leetcode]Construct Binary Tree from Inorder and Postorder Traversal @ Python

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