python实现剑指offer系列:重建二叉树

xiaoxiao2021-02-28  165

题目:输入某二叉树的前序遍历和中序遍历的结果,请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6},则重建二叉树并返回。

代码:

class Node: def __init__(self, data, left, right): self.data = data self.left = left self.right = right def construct_tree(pre_order, mid_order): # if len(pre_order) == 0 or len(mid_oder) ==0 : return None # 前序遍历的第一个结点一定是根结点 root_data = pre_order[0] for i in range(0, len(mid_order)): if mid_order[i] == root_data: break # 递归构造左子树和右子树 left = construct_tree(pre_order[1 : 1 + i], mid_order[:i]) right = construct_tree(pre_order[1 + i:], mid_order[i+1:]) return Node(root_data, left, right) if __name__ == '__main__': pre_order = [1, 2, 4, 7, 3, 5, 6, 8] mid_order = [4, 7, 2, 1, 5, 3, 8, 6] root = construct_tree(pre_order, mid_order)
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