二叉树(前序，中序，后序，层序)遍历递归与循环的python实现（五颗星）
113. Path Sum II
Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
5
/
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
Return:
[
[5,4,11,2],
[5,8,4,5]
]
- 思路：利用递归的思想解决
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def pathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: List[List[int]]
"""
import copy
def find_sum_tree(root, he, sum, item, result):
if root == None:
return
item.append(root.val)
he += root.val
if he == sum and not root.left and not root.right:
item_deep = copy.deepcopy(item)
result.append(item_deep)
find_sum_tree(root.left, he, sum, item, result)
find_sum_tree(root.right, he, sum, item, result)
he -= item[-1]
item.pop()
he = 0
item = []
result = []
find_sum_tree(root, he, sum, item, result)
return result
题目描述
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。
- python 解答
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回构造的TreeNode根节点
def reConstructBinaryTree(self, pre, tin):
# write code here
if not pre or not tin:
return None
node = TreeNode(pre.pop(0))
index = tin.index(node.val)
node.left = self.reConstructBinaryTree(pre, tin[ :index])
node.right = self.reConstructBinaryTree(pre, tin[index+1: ])
return node
98. 验证二叉搜索树
给定一个二叉树，判断其是否是一个有效的二叉搜索树。
假设一个二叉搜索树具有如下特征：
节点的左子树只包含小于当前节点的数。
节点的右子树只包含大于当前节点的数。
所有左子树和右子树自身必须也是二叉搜索树。
示例 1:
输入:
2
/
1 3
输出: true
示例 2:
输入:
5
/
1 4
/
3 6
输出: false
解释: 输入为: [5,1,4,null,null,3,6]。
根节点的值为 5 ，但是其右子节点值为 4 。
- python 解答一：
二叉搜索树的中序遍历一定是一个递增的序列，先得到中序遍历的list,之后再进行判断是否是递增的。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
def inorder(root):
if root == None:
return []
return inorder(root.left) + [root.val] + inorder(root.right)
inorder_list = inorder(root)
if inorder_list == sorted(list(set(inorder_list))):
return True
return False
- 思路二：
利用递归的思想，判断是否所有的节点是否都是满足条件。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
'''
def inorder(root):
if root == None:
return []
return inorder(root.left) + [root.val] + inorder(root.right)
inorder_list = inorder(root)
if inorder_list == sorted(list(set(inorder_list))):
return True
return False
'''
return self.helper(root)
def helper(self, root, left=None, right=None):
if not root:
return True
if left and left.val >= root.val:
return False
if right and right.val <= root.val:
return False
return self.helper(root.left, left, root) and\
self.helper(root.right, root, right)
给定一个 N 叉树，返回其节点值的前序遍历。
例如，给定一个 3叉树 :
返回其前序遍历: [1,3,5,6,2,4]。
- python 解答
递归方法
"""
# Definition for a Node.
class Node(object):
def __init__(self, val, children):
self.val = val
self.children = children
"""
class Solution(object):
def preorder(self, root):
"""
:type root: Node
:rtype: List[int]
"""
results = []
self.function(root, results)
return results
def function(self, root, results):
if root is None:
return
results.append(root.val)
for child in root.children:
if child is not None:
self.function(child, results)