树的简单算法题

二叉树插入

class TreeNode:
    def __init__(self, val, left, right):
        self.val = val
        self.left = left
        self.right = right
      
class BinarySearchTree:
    def insert(self, root, val):
        if root == None:
            root = TreeNode(val, None, None)
        else:
            if val < root.val:
                root.left = self.insert(root.left, val)
            if val > root.val:
                root.right = self.insert(root.right, val)
        return root    

    def preOrder(self, root):
        if root:
            print root.val
            self.preOrder(root.left)
            self.preOrder(root.right)

Tree = BinarySearchTree()
root = None
for i in [1,2,3]:
    root = Tree.insert(root, i)
Tree.preOrder(root)

有序数组创建二叉树

class Solution:  
    def sortedArrayToBST(self, num):  
        if not num:  
            return None  
          
        mid = len(num)//2  #“//”表示整数除法；“/”浮点数除法；  
        root = TreeNode(num[mid])  
        left = num[:mid]  
        right = num[mid+1:]  
        root.left = self.sortedArrayToBST(left)  
        root.right = self.sortedArrayToBST(right)  
        return root  

遍历二叉树

algorithms/

前序 根左右

中序 左根右

后序 左右根

递归方式

class Tree(object):
    def __init__(self,data,left,right):
        self.data=data
        self.left=left
        self.right=right
def post_visit(Tree):
    if Tree:
        post_visit(Tree.left)
        post_visit(Tree.right)
        print Tree.data
def pre_visit(Tree):
    if Tree:
        print Tree.data
        pre_visit(Tree.left)
        pre_visit(Tree.right)
def in_visit(Tree):
    if Tree:
        in_visit(Tree.left)
        print Tree.data
        in_visit(Tree.right)

非递归

class TreeNode:  
    def __init__(self,value=None,leftNode=None,rightNode=None):  
        self.value = value  
        self.leftNode = leftNode  
        self.rightNode = rightNode  
  
  
class Tree:  
    def __init__(self,root=None):  
        self.root = root  
  
    def preOrder(self):  
        if not self.root:  
            return  
        stackNode = []  
        stackNode.append(self.root)  
        while stackNode:  
            node = stackNode.pop()  
            print node.value,  
            if node.rightNode:  
                stackNode.append(node.rightNode)  
            if node.leftNode:  
                stackNode.append(node.leftNode)  

平衡二叉树

输入一棵二叉树，判断该二叉树是否是平衡二叉树。

class Solution:
    def getDepth(self , Root):
        if Root == None:
            return 0;
        lDepth = self.getDepth(Root.left);
        rDepth = self.getDepth(Root.right);
        return max(lDepth , rDepth) + 1;

    def IsBalanced_Solution(self, pRoot):
        if not pRoot:
            return True
        lDepth = self.getDepth(pRoot.left);
        rDepth = self.getDepth(pRoot.right);
        diff = lDepth - rDepth;
        if diff < -1 or diff > 1:
            return False;
        return self.IsBalanced_Solution(pRoot.left) and self.IsBalanced_Solution(pRoot.right);

判断一棵树是否为另一棵的子树

给定两棵二叉树，判断T2是否是T1某棵子数的结构。

T1序列化成字符串str1;

T2序列化成字符串str2;

用KMP算法判断str1中是否包含str2:如果str1包含str2,说明T1包含于T2结构一致的子树。

KMP算法

#KMP  
def kmp_match(s, p):  
    m = len(s); n = len(p)  
    cur = 0#起始指针cur  
    table = partial_table(p)  
    while cur<=m-n:  
        for i in range(n):  
            if s[i+cur]!=p[i]:  
                cur += max(i - table[i-1], 1)#有了部分匹配表,我们不只是单纯的1位1位往右移,可以一次移动多位  
                break  
        else:  
            return True  
    return False  
  
#部分匹配表  
def partial_table(p):  
    '''''partial_table("ABCDABD") -> [0, 0, 0, 0, 1, 2, 0]'''  
    prefix = set()  
    postfix = set()  
    ret = [0]  
    for i in range(1,len(p)):  
        prefix.add(p[:i])  
        postfix = {p[j:i+1] for j in range(1,i+1)}  
        ret.append(len((prefix&postfix or {''}).pop()))  
    return ret