题目:
题目链接: https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
解题思路:
前序遍历:根节点 -> 左子树 -> 右子树
中序遍历:左子树 -> 根节点 -> 右子树
后序遍历:左子树 -> 右子树 -> 根节点
根据遍历规则,在知道前序遍历的第一个值就是当前的根节点
因为题目中说明没有重复值,所以使用根节点的值,将中序遍历分为左右两部分,左边的值为左子树,右边的值为右子树
重复此步骤,直到所有元素都用完为止
代码实现:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
def build_help(preorder, inorder):
if 0 == len(preorder) or 0 == len(inorder):
return None
in_index = inorder.index(preorder[0])
if in_index == -1:
return None
node = TreeNode(preorder[0])
del preorder[0]
del inorder[in_index]
node.left = build_help(preorder, inorder[:in_index])
node.right = build_help(preorder, inorder[in_index:])
return node
return build_help(preorder, inorder)