通过前序遍历和中序遍历构建二叉树 python实现
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通过前序遍历和中序遍历构建二叉树 python实现
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前言
通過前序遍歷和中序遍歷構建二叉樹的原理,主要是找前序遍歷根節點在中序遍歷中的位置,然后將二叉樹而成左子樹和右子樹,然后依次進行這樣的操作,思路還是比較簡單的
代碼
class Node:def __init__(self, data, left, right):self.data = dataself.left = leftself.right = rightdef construct_tree(pre_order, mid_order):# 忽略參數合法性判斷if len(pre_order) == 0:return None# 前序遍歷的第一個結點一定是根結點root_data = pre_order[0]i = mid_order.index(root_data)# 遞歸構造左子樹和右子樹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)
如果加上遍歷
這里以層次遍歷來遍歷二叉樹
# coding:utf-8# !/usr/bin/env python# Time: 2018/6/6 9:02# Author: sty# File: construct_tree.pyclass Node:def __init__(self, data, left, right):self.data = dataself.left = leftself.right = rightdef construct_tree(pre_order, mid_order):# 忽略參數合法性判斷if len(pre_order) == 0:return None# 前序遍歷的第一個結點一定是根結點root_data = pre_order[0]i = mid_order.index(root_data)# 遞歸構造左子樹和右子樹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)def levelOrder(root):"""普通二叉樹層次遍歷,并輸出:type root: TreeNode:rtype: List[List[int]]"""res, level = [], [root]while root and level:currentNode = []nextLevel = []for node in level:currentNode.append(node.data)if node.left:nextLevel.append(node.left)if node.right:nextLevel.append(node.right)res.append(currentNode)level = nextLevelreturn resif __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)print(levelOrder(root))print root.dataprint root.left.dataprint root.right.dataprint root.left.left.dataprint root.right.left.dataprint root.right.right.dataprint root.left.left.right.dataprint root.right.right.left.data
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