【LeetCode】106. Construct Binary Tree from Inorder and Postorder Traversal 解题报告
【LeetCode】106. Construct Binary Tree from Inorder and Postorder Traversal 解题报告(Python)
标签: LeetCode
题目描述:
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
题目大意
根据中序遍历和后序遍历重建二叉树。
解题方法
这个是套路题,我没有完全记住每一步是多少,而是根据树的样子和两个数组进行分析,得出切片的位置。
后序遍历的最后一个元素一定是根节点,在中序遍历中找出此根节点的位置序号。中序遍历序号左边的是左孩子,右边的是右孩子。再根据左孩子和右孩子的长度对后序遍历进行切片即可。
# 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 or not postorder: return None
val = postorder[-1]
root = TreeNode(val)
index = inorder.index(val)
root.left = self.buildTree(inorder[:index], postorder[:index])
root.right = self.buildTree(inorder[index+1:], postorder[index:-1])
return root
日期
2018 年 3 月 12 日
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