106. 从中序与后序遍历序列构造二叉树

根据一棵树的中序遍历与后序遍历构造二叉树。

注意: 你可以假设树中没有重复的元素。

例如，给出

中序遍历 inorder = [9,3,15,20,7]后序遍历 postorder = [9,15,7,20,3]返回如下的二叉树：    3   / \  9  20    /  \   15   7

来源：力扣(LeetCode)

链接：https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

Link：https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

递归

基本的思路是用postorder最后一个作为跟节点，然后根据inorder值的位置，来确定左子树+右子树

普通递归

class Solution:    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:        if len(inorder) == 0 or len(postorder) == 0:            return None        val = postorder.pop()        root = TreeNode(val)        i = inorder.index(val)        root.right = self.buildTree(inorder[i + 1:], postorder)        root.left = self.buildTree(inorder[:i], postorder)        return root

优化递归

⚠️注意，由于没有重复元素

1.先用字典做一个反向索引，加速查找index的速度2.使用下标，而不是新建一个数组

    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:        self.map_inorder = {}        for i in range(len(inorder)):            self.map_inorder[inorder[i]] = i        return self.helper(inorder, postorder, 0, len(inorder) - 1)    def helper(self, inorder: List[int], postorder: List[int], low: int, high: int) -> TreeNode:        if low > high:            return None        val = postorder.pop()        root = TreeNode(val)        i = self.map_inorder[val]        root.right = self.helper(inorder, postorder, i + 1, high)        root.left = self.helper(inorder, postorder, low, i - 1)        return root

空间优化递归

先创建右子树部分，然后在做左边

class Solution:    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:        return self.helper(inorder, postorder, None)    def helper(self, inorder: List[int], postorder: List[int], stop: TreeNode) -> TreeNode:        if len(inorder) == 0 or inorder[-1] == stop:            return None        val = postorder.pop()        root = TreeNode(val)        root.right = self.helper(inorder, postorder, val)        inorder.pop()        root.left = self.helper(inorder, postorder, stop)        return root

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