[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:
# 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
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