算法编程 dfs 从先序和中序遍历还原二叉树

105. 从前序与中序遍历序列构造二叉树

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

 

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]

示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

 

提示:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder 和 inorder 均 无重复 元素
• inorder 均出现在 preorder
• preorder 保证 为二叉树的前序遍历序列
• inorder 保证 为二叉树的中序遍历序列

 

# Definition for a binary tree node.
# 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, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        inorder_map = {val:i for i,val in enumerate(inorder)}
        n = len(preorder)
        nth = 0

        def dfs(start, end):
            nonlocal nth
            if start > end:
                return None

            val = preorder[nth]
            i = inorder_map[val] 

            root = TreeNode(val)
            if start <= i-1:
                nth += 1
                root.left = dfs(start, i-1)                            
            if i+1 <= end:
                nth += 1                        
                root.right = dfs(i+1, end)

            return root
        
        return dfs(0, n-1)