Time limit: 5000MS
Memory limit: 589824KB
Title description:
Given two integer arrays inorder and postorder, where inorder is the inorder traversal of a binary tree, and postorder is the postorder traversal of the same tree, please return the preorder traversal of this binary tree.
Input description
The in-order traversal of the input element binary tree and the post-order traversal of the same tree
Output description
The output is the pre-order traversal of the binary tree
Sample input
9 3 15 20 7
9 15 7 20 3
Sample output
3 9 20 15 7
"""
算法思路：根据中序遍历和后序遍历重建二叉树，然后先序遍历即可。
具体步骤如下：
1. 后序遍历的最后一个节点即为根节点，找到根节点在中序遍历中的位置，将中序遍历分为左子树和右子树。
2. 根据左子树和右子树的长度，将后序遍历也分为左子树和右子树。
3. 递归重建左子树和右子树。
4. 先序遍历的顺序为根节点、左子树、右子树。
时间复杂度：$O(n)$，其中$n$为二叉树节点个数。
空间复杂度：$O(n)$，需要存储中序遍历和后序遍历的数组，以及递归时的栈空间。
参考代码：
如下
"""
from typing import List
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
if not inorder or not postorder:
return None
root_val = postorder[-1]
root = TreeNode(root_val)
idx = inorder.index(root_val)
root.left = self.buildTree(inorder[:idx], postorder[:idx])
root.right = self.buildTree(inorder[idx+1:], postorder[idx:-1])
return root
def preorderTraversal(self, root: TreeNode) -> List[int]:
if not root:
return []
res = []
stack = [root]
while stack:
node = stack.pop()
res.append(node.val)
if node.right:
stack.append(node.right)
if node.left:
stack.append(node.left)
return res
if __name__ == '__main__':
inorder = list(map(int, input().split()))
postorder = list(map(int, input().split()))
s = Solution()
root = s.buildTree(inorder, postorder)
res = s.preorderTraversal(root)
print(' '.join(map(str, res)))