Java 给定一棵二叉树,找出同一级别上两个节点之间的水平距离,同时计算节点不存在的位置
让我非常清楚,am没有得到水平距离是多少 但还是从我的角度来看。水平距离是指:在同一级别的给定节点之间缺少或存在节点 在我的例子中,当我试图找出Java 给定一棵二叉树,找出同一级别上两个节点之间的水平距离,同时计算节点不存在的位置,java,algorithm,data-structures,tree,binary-tree,Java,Algorithm,Data Structures,Tree,Binary Tree,让我非常清楚,am没有得到水平距离是多少 但还是从我的角度来看。水平距离是指:在同一级别的给定节点之间缺少或存在节点 在我的例子中,当我试图找出7和1之间的距离时,我得到了输出,即2。这就是为什么我会这样想 但是如果我试图找出9和6之间的距离,我得到的输出是4 例如,在给定树中,处于同一级别的节点7和1之间的距离为2 (考虑节点2的右子节点和节点3的左子节点) 这张图片将帮助你理解 下面是我用来检查距离的代码 public class BinaryHorizontalDistance {
7
和1
之间的距离时,我得到了输出,即2。这就是为什么我会这样想
但是如果我试图找出9
和6
之间的距离,我得到的输出是4
例如,在给定树中,处于同一级别的节点7和1之间的距离为2
(考虑节点2的右子节点和节点3的左子节点)
这张图片将帮助你理解
下面是我用来检查距离的代码
public class BinaryHorizontalDistance
{
public int findDistance(Node root, int n1, int n2)
{
int leftNodeToRootNode = Pathlength(root, n1, "leftNodeToRootNode") - 2;
int rightNodeToRootNode = Pathlength(root, n2,"rightNodeToRootNode") - 2;
int lcaData = findLCA(root, n1, n2).data; //LCA->Lowest Common Ancestor
int lcaDistance = Pathlength(root, lcaData,"lcaDistance") - 1;
return (leftNodeToRootNode + rightNodeToRootNode) - 2 * lcaDistance;
}
public int Pathlength(Node root, int n1,String callingFrom)
{
if (root != null)
{
int x = 0;
if("rightNodeToRootNode" == callingFrom)
{
if(root.left ==null && root.right ==null)
{
//do nothing
}
else if(root.left ==null || root.right ==null)
{
System.out.println("counting the position where the node is not present is : " + root.data);
}
if ((root.data == n1) || (x = Pathlength(root.left,
n1,"rightNodeToRootNode")) > 0 || (x = Pathlength(root.right,
n1,"rightNodeToRootNode")) > 0)
{
return x + 1;
}
}
if("rightNodeToRootNode" != callingFrom )
{
if ((root.data == n1) || (x = Pathlength(root.left,
n1,"leftNodeToRootNode")) > 0 || (x = Pathlength(root.right,
n1,"leftNodeToRootNode")) > 0)
{
return x + 1;
}
}
return 0;
}
return 0;
}
public Node findLCA(Node root, int n1, int n2)
{
if (root != null)
{
if (root.data == n1 || root.data == n2)
{
return root;
}
Node left = findLCA(root.left, n1, n2);
Node right = findLCA(root.right, n1, n2);
if (left != null && right != null)
{
return root;
}
if (left != null)
{
return left;
}
if (right != null)
{
return right;
}
}
return null;
}
public static void main(String[] args) throws java.lang.Exception
{
Node root = new Node(5);
root.right = new Node(2);
root.left = new Node(3);
root.right.right = new Node(7);
//root.right.left = new Node(78);
root.right.right.right = new Node(9);
root.left.left = new Node(1);
//root.left.right = new Node(22);
root.left.left.right = new Node(4);
root.left.left.left = new Node(6);
BinaryHorizontalDistance binaryTreeTest = new BinaryHorizontalDistance();
System.out.println("Distance between 7 and 1 is : " +
binaryTreeTest.findDistance(root,9, 6));
}
}
class Node
{
int data;
Node left;
Node right;
public Node(int data)
{
this.data = data;
this.left = null;
this.right = null;
}
}
请举例说明。很高兴进一步解释您知道以下定义:
- 如果你是一个左撇子:计数-1
- 如果您是正确的孩子:计数+1
h(4,6)
- 4是5岁以下的孩子:-1
- 6是5的正确孩子:+1
h(2,6)
,2
是右4的子节点(**显然,如果节点是唯一的子节点,则必须将其视为右子节点):
所以h(2,4)=+1
recall
h(4,6)=0
所以h(2,6)=1
关于你的一个例子,说h(9,6)
**我认为选择+1是为了保持一致性,但我只是观察了Python中的代码
"""
Given a binary tree, find the horizontal distance between 2 nodes at the same level, also counting the position where the node is not present.
"""
# binary tree node
class Node:
# Constructor to create new node
def __init__(self, data):
self.data = data
self.left = self.right = None
self.val = data
# This function returns pointer to LCA of two given values n1 and n2.
def LCA(root, n1, n2):
# Base case
if root is None:
return None
# If either n1 or n2 matches with root's
# key, report the presence by returning
# root
if root.data == n1 or root.data == n2:
return root
if root.data == None or root.data == None:
return None
# Look for keys in left and right subtrees
left = LCA(root.left, n1, n2)
right = LCA(root.right, n1, n2)
if left is not None and right is not None:
return root
# Otherwise check if left subtree or
# right subtree is LCA
if left:
return left
else:
return right
# function to find distance of any node
# from root
def findLevel(root, data, d, level):
# Base case when tree is empty
if root is None:
return
# Node is found then append level
# value to list and return
if root.data == data:
d.append(level)
return
findLevel(root.left, data, d, level + 1)
findLevel(root.right, data, d, level + 1)
# function to find distance between two
# nodes in a binary tree
def findDistance(root, n1, n2):
lca = LCA(root, n1, n2)
# to store distance of n1 from lca
d1 = []
# to store distance of n2 from lca
d2 = []
# if lca exist
if lca:
# distance of n1 from lca
findLevel(lca, n1, d1, 0)
# print(d1)
# distance of n2 from lca
findLevel(lca, n2, d2, 0)
# print(d2)
return d1[0]
else:
return -1
def inorder(root):
if root:
# Traverse left
inorder(root.left)
# Traverse root
ls.append(root.val)
# print(str(root.val) + "->", end='')
# Traverse right
inorder(root.right)
def height(root):
if root:
return 1+max(height(root.left), height(root.right))
else:
return -1
# Driver program to test above function
root = Node(5)
root1 = root.val
root.left = Node(3)
root.right = Node(2)
root.left.left = Node(1)
root.left.left.left = Node(6)
root.left.left.right = Node(4)
root.right.right= Node(7)
root.right.right.right= Node(9)
# print("Height of the Binary Tree: ", height(root))
treeHeight = height(root)
# Total nodes required to compelete binary tree
totalNodes = 2**(treeHeight + 1) -1
# print("Required Nodes : ",2**(treeHeight + 1) -1)
ls =[]
# print("Inorder traversal ")
inorder(root)
index = ls.index(root1)
treeLeft =[]
treeRight = []
for i in range(len(ls)):
if i < index:
treeLeft.append(ls[i])
elif i == index:
pass
else:
treeRight.append(ls[i])
print("Nodes at Same Level and horizontal distance between 2 nodes ")
print()
print("---**---**---**---**---**---**---**---**---**---**---**---**---**---")
print()
for i in treeLeft:
for j in treeRight:
# print("Dist(",i,root1,") = ", findDistance(root, i, root1))
# print("Dist(",root1,j,") = ", findDistance(root, j, root1))
if findDistance(root, i, root1) == findDistance(root, j, root1):
print("Nodes are : (",i,",",j, ") & Horizontal Distance between (",i,",",j,"): ", findDistance(root, i, root1))
print()
print("---**---**---**---**---**---**---**---**---**---**---**---**---**---")
print()
print("count of the position where the node is not present : ", totalNodes - len(ls))
print()
print("---**---**---**---**---**---**---**---**---**---**---**---**---**---")
"""
Program Output :
Nodes at Same Level and horizontal distance between 2 nodes
---**---**---**---**---**---**---**---**---**---**---**---**---**---
Nodes are : ( 6 , 9 ) & Horizontal Distance between ( 6 , 9 ): 3
Nodes are : ( 1 , 7 ) & Horizontal Distance between ( 1 , 7 ): 2
Nodes are : ( 4 , 9 ) & Horizontal Distance between ( 4 , 9 ): 3
Nodes are : ( 3 , 2 ) & Horizontal Distance between ( 3 , 2 ): 1
---**---**---**---**---**---**---**---**---**---**---**---**---**---
count of the position where the node is not present : 7
---**---**---**---**---**---**---**---**---**---**---**---**---**---
"""
“”“
给定一棵二叉树,找出同一级别上两个节点之间的水平距离,同时计算节点不存在的位置。
"""
#二叉树节点
类节点:
#用于创建新节点的构造函数
定义初始化(自身,数据):
self.data=数据
self.left=self.right=无
self.val=数据
#此函数返回指向两个给定值n1和n2的LCA的指针。
def LCA(根、n1、n2):
#基本情况
如果root为None:
一无所获
#如果n1或n2与根匹配
#键,通过返回来报告存在
#根
如果root.data==n1或root.data==n2:
返回根
如果root.data==无或root.data==无:
一无所获
#在左子树和右子树中查找关键点
左=LCA(根左,n1,n2)
右=LCA(根,右,n1,n2)
如果“左”不是“无”,而“右”不是“无”:
返回根
#否则,请检查左子树或
#右子树是LCA
如果留下:
左转
其他:
返回权
#函数查找任意节点的距离
#从根本上
def findLevel(根、数据、d、级别):
#树为空时的基本情况
如果root为None:
返回
#找到节点,然后追加级别
#要列出并返回的值
如果root.data==数据:
d、 附加(级别)
返回
findLevel(root.left,数据,d,级别+1)
findLevel(root.right,数据,d,级别+1)
#函数查找两个对象之间的距离
#二叉树中的节点
def FindInstance(根、n1、n2):
lca=lca(根,n1,n2)
#存储n1与lca之间的距离
d1=[]
#储存氮气与lca之间的距离
d2=[]
#如果存在生命周期评价
如果是生命周期评价:
#n1与lca的距离
findLevel(生命周期评价,n1,d1,0)
#印刷品(d1)
#n2与lca的距离
findLevel(生命周期评价,n2,d2,0)
#印刷品(d2)
返回d1[0]
其他:
返回-1
def索引(根):
如果根:
#向左移动
顺序(根.左)
#横移根
ls.append(root.val)
#打印(str(root.val)+“->”,end='')
#导线权
顺序(root.right)
def高度(根):
如果根:
返回1+最大值(高度(根左),高度(根右))
其他:
返回-1
#用于测试上述功能的驱动程序
根=节点(5)
root1=root.val
root.left=节点(3)
root.right=节点(2)
root.left.left=节点(1)
root.left.left.left=节点(6)
root.left.left.right=节点(4)
root.right.right=节点(7)
root.right.right.right=节点(9)
#打印(“二叉树的高度:,高度(根))
树高=高度(根)
#完成二叉树所需的总节点数
totalNodes=2**(树高+1)-1
#打印(“必需节点:”,2**(树高+1)-1)
ls=[]
#打印(“按顺序遍历”)
顺序(根)
index=ls.index(root1)
treeLeft=[]
treeRight=[]
对于范围内的i(len(ls)):
如果i<索引:
treeLeft.append(ls[i])
elif i==索引:
通过
其他:
treerRight.append(ls[i])
打印(“同一级别的节点和两个节点之间的水平距离”)
打印()
打印(“--***--***--***--***--***--***--***--***--***--***--***--***--***--***--***--***--***--”)
打印()
对于我在treeLeft:
对于树右中的j:
#打印(“Dist(“,i,root1,”)=”,FindInstance(root,i,root1))
#打印(“Dist(“,root1,j,”)=”,findInstance(root,j,root1))
如果FindInstance(root,i,root1)=FindInstance(root,j,root1):
打印(“节点是:(“,i,”,“,j)”)和(“,i,”,“,j,”):”,FindInstance(根,i,根1))之间的水平距离
打印()
打印(“--***--***--***--***--***--***--***--***--***--***--***--***--***--***--***--***--***--”)
打印()
打印(“节点不存在的位置计数:”,totalNodes-len(ls))
打印()
打印(“--***--***--***--***--***--***--***--***--***--***--***--***--***--***--***--***--***--”)
"""
程序输出:
同一级别的节点和两个节点之间的水平距离
---**---**---**---**---**---**---**---**---**---**---**---**---**---
节点是:(6,9)&和(6,9)之间的水平距离:3
节点是:(1,7)&和(1,7)之间的水平距离:2
节点是:(4,9)和水平线
5
/ \
4 6
\
2
h(9,7) = 2
h(2,3) = 0
h(3,1) = 1 (1 only child so +1)
h(1,6) = 1 (same)
total: 4
"""
Given a binary tree, find the horizontal distance between 2 nodes at the same level, also counting the position where the node is not present.
"""
# binary tree node
class Node:
# Constructor to create new node
def __init__(self, data):
self.data = data
self.left = self.right = None
self.val = data
# This function returns pointer to LCA of two given values n1 and n2.
def LCA(root, n1, n2):
# Base case
if root is None:
return None
# If either n1 or n2 matches with root's
# key, report the presence by returning
# root
if root.data == n1 or root.data == n2:
return root
if root.data == None or root.data == None:
return None
# Look for keys in left and right subtrees
left = LCA(root.left, n1, n2)
right = LCA(root.right, n1, n2)
if left is not None and right is not None:
return root
# Otherwise check if left subtree or
# right subtree is LCA
if left:
return left
else:
return right
# function to find distance of any node
# from root
def findLevel(root, data, d, level):
# Base case when tree is empty
if root is None:
return
# Node is found then append level
# value to list and return
if root.data == data:
d.append(level)
return
findLevel(root.left, data, d, level + 1)
findLevel(root.right, data, d, level + 1)
# function to find distance between two
# nodes in a binary tree
def findDistance(root, n1, n2):
lca = LCA(root, n1, n2)
# to store distance of n1 from lca
d1 = []
# to store distance of n2 from lca
d2 = []
# if lca exist
if lca:
# distance of n1 from lca
findLevel(lca, n1, d1, 0)
# print(d1)
# distance of n2 from lca
findLevel(lca, n2, d2, 0)
# print(d2)
return d1[0]
else:
return -1
def inorder(root):
if root:
# Traverse left
inorder(root.left)
# Traverse root
ls.append(root.val)
# print(str(root.val) + "->", end='')
# Traverse right
inorder(root.right)
def height(root):
if root:
return 1+max(height(root.left), height(root.right))
else:
return -1
# Driver program to test above function
root = Node(5)
root1 = root.val
root.left = Node(3)
root.right = Node(2)
root.left.left = Node(1)
root.left.left.left = Node(6)
root.left.left.right = Node(4)
root.right.right= Node(7)
root.right.right.right= Node(9)
# print("Height of the Binary Tree: ", height(root))
treeHeight = height(root)
# Total nodes required to compelete binary tree
totalNodes = 2**(treeHeight + 1) -1
# print("Required Nodes : ",2**(treeHeight + 1) -1)
ls =[]
# print("Inorder traversal ")
inorder(root)
index = ls.index(root1)
treeLeft =[]
treeRight = []
for i in range(len(ls)):
if i < index:
treeLeft.append(ls[i])
elif i == index:
pass
else:
treeRight.append(ls[i])
print("Nodes at Same Level and horizontal distance between 2 nodes ")
print()
print("---**---**---**---**---**---**---**---**---**---**---**---**---**---")
print()
for i in treeLeft:
for j in treeRight:
# print("Dist(",i,root1,") = ", findDistance(root, i, root1))
# print("Dist(",root1,j,") = ", findDistance(root, j, root1))
if findDistance(root, i, root1) == findDistance(root, j, root1):
print("Nodes are : (",i,",",j, ") & Horizontal Distance between (",i,",",j,"): ", findDistance(root, i, root1))
print()
print("---**---**---**---**---**---**---**---**---**---**---**---**---**---")
print()
print("count of the position where the node is not present : ", totalNodes - len(ls))
print()
print("---**---**---**---**---**---**---**---**---**---**---**---**---**---")
"""
Program Output :
Nodes at Same Level and horizontal distance between 2 nodes
---**---**---**---**---**---**---**---**---**---**---**---**---**---
Nodes are : ( 6 , 9 ) & Horizontal Distance between ( 6 , 9 ): 3
Nodes are : ( 1 , 7 ) & Horizontal Distance between ( 1 , 7 ): 2
Nodes are : ( 4 , 9 ) & Horizontal Distance between ( 4 , 9 ): 3
Nodes are : ( 3 , 2 ) & Horizontal Distance between ( 3 , 2 ): 1
---**---**---**---**---**---**---**---**---**---**---**---**---**---
count of the position where the node is not present : 7
---**---**---**---**---**---**---**---**---**---**---**---**---**---
"""