《剑指offer》【二叉搜索树的后序遍历序列】

最新推荐文章于 2024-05-27 21:18:06 发布

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最新推荐文章于 2024-05-27 21:18:06 发布

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本文介绍了一种算法，用于验证一个整数序列是否可能作为某棵二叉搜索树的后序遍历结果。通过递归地将序列分为左子树和右子树，检查每个子树是否满足二叉搜索树的性质。

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class Solution:
    def VerifySquenceOfBST(self, sequence):
        # write code here
        if len(sequence) == 0:
            return False
        else:
            root = sequence[-1]
            del sequence[-1]
            lefttree = []
            righttree =[]
            # 左子树和右子树分界
            splitindex = -1
            for i in range(len(sequence)):
                # 值小于根结点的归为左子树
                if sequence[i] < root:
                    lefttree.append(sequence[i])
                    splitindex = i
                else:
                    break
            for i in range(splitindex+1, len(sequence)):
                # 若右子树部分有小于根结点的值，说明不是二叉搜索树
                if sequence[i] > root:
                    righttree.append(sequence[i])
                else:
                    return False
            if len(lefttree) <= 1:
                left = True
            else:
                # 递归判断左子树
                left = self.VerifySquenceOfBST(lefttree)
            if len(righttree) <= 1:
                right = True
            else:
                right = self.VerifySquenceOfBST(righttree)
            return left and right