给定一棵二叉树，求同一级别的 2 个节点之间的水平距离，同时计算该节点不存在的 position

Question

让我很清楚，我没有得到水平距离是什么。

但从我的角度来看。 水平距离是指：同一级别的给定节点之间缺少或存在的节点。

在我的情况下，当我试图找出早上7点和1点之间的距离时，得到 output 即 2。这就是为什么我想上面提到的方式。

但是，如果我尝试找出上午9点和6点之间的距离，则将 output 设为 4。

例如，在给定的树中，同一级别的节点 7 和 1 之间的距离为 2（考虑节点 2 的右孩子和节点 3 的左孩子）

这张图可以帮助你理解

以下是用于检查距离的代码。

    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;
    }
}

一个例子的解释将不胜感激。 并很高兴进一步解释

Answer 1

你知道定义：

• 如果您是左孩子：计数-1，
• 如果您是正确的孩子：计数+1

        5
      /   \
     4     6

在这里，计算h(4,6)

• 4 是 5 的左孩子：-1
• 6 是 5 的右孩子：+1

h(4,6) = 0

        5
      /   \
     4     6
      \ 
       2

在这里，要计算h(2,6) ， 2是 4 的右孩子（**显然，如果 node 是唯一的孩子，它必须被视为右孩子）：

所以h(2,4)=+1召回h(4,6)=0所以h(2,6) = 1

关于你的一个例子，比如h(9,6)

h(9,7) = 2
h(2,3) = 0
h(3,1) = 1 (1 only child so +1)
h(1,6) = 1 (same)
total: 4

** 我想选择 +1 是为了保持一致性，但我只是观察到了它

Answer 2

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

---**---**---**---**---**---**---**---**---**---**---**---**---**---
"""