力扣105. 从前序与中序遍历序列构造二叉树

原题105

class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
        def helper(prebegin,preend,inbegin,inend):
            if prebegin > preend or inbegin > inend:return None
            root = TreeNode(preorder[prebegin])
            i = 0
            while i + inbegin < inend:
                if inorder[i + inbegin] == preorder[prebegin]:
                    break
                i += 1
            root.left = helper(prebegin+1,prebegin+i,inbegin,inbegin + i - 1)
            root.right = helper(prebegin + i + 1,preend,inbegin + i + 1,inend)
            return root
        return helper(0,len(preorder)-1,0,len(inorder)-1);

 原题106

class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
        def helper(postbegin,postend,inbegin,inend):
            if postbegin > postend or inbegin > inend:return None
            root = TreeNode(postorder[postend])
            i = 0
            while i + inbegin < inend:
                if inorder[i + inbegin] == postorder[postend]:
                    break
                i += 1
            root.left = helper(postbegin,postbegin+i-1,inbegin,inbegin + i - 1)
            root.right = helper(postbegin + i,postend-1,inbegin + i + 1,inend)
            return root
        return helper(0,len(postorder)-1,0,len(inorder)-1);