给定一棵二叉树和一个整数K ，任务是计算给定二叉树中可能的叶节点对，使得它们之间的距离至多为K 。
例子：
Input: K = 3
Output: 1
Explanation:
The leaf nodes of the tree are 3 and 4
And the shortest distance between them is 3.
This is the only valid pair.
Input: K = 3
Output: 2
方法：这个想法是使用大小为K + 1的数组的后序遍历来跟踪特定距离的节点数。以下是步骤：
- 初始化大小为K + 1的数组arr[] ，其中arr[i]表示距离当前节点i处的叶节点数。
- 然后，分别递归更新数组left[]和right[]中每个节点的左子树和右子树的上述数组。
- 在上述步骤之后，对于每个节点，对应索引处的 left[] 和 right[] 之间的距离将给出最左侧和最右侧叶节点之间的距离。在数组res[]中将其更新为：
res[i + 1] = left[i] * right[i]
- 现在对于数组left[]和right[] 中所有可能的对(l, r) ，如果它们之间的距离之和最多为 K ，则将对的计数增加为：
count += left[l] * right[r]
下面是上述方法的实现：
C++14
1
/ \
2 3
/
4
Java
1
/ \
2 3
/ \ / \
4 5 6 7
Python3
// C++14 implementation of the
// above approach
#include
using namespace std;
// Structure of a Node
struct Node
{
int data;
Node* left, *right;
// Constructor of the class
Node(int item)
{
data = item;
left = right = NULL;
}
};
// Stores the count of required pairs
int result = 0;
// Function to perform dfs to find pair of
// leaf nodes at most K distance apart
vector dfs(Node* root, int distance)
{
// Return empty array if node is NULL
if (root == NULL)
{
vector res(distance + 1, 0);
return res;
}
// If node is a leaf node and return res
if (root->left == NULL &&
root->right == NULL)
{
vector res(distance + 1, 0);
res[1]++;
return res;
}
// Traverse to the left
vector left = dfs(root->left,
distance);
// Traverse to the right
vector right = dfs(root->right,
distance);
vector res(distance + 1, 0);
// Update the distance between left
// and right leaf node
for(int i = res.size() - 2;
i >= 1; i--)
res[i + 1] = left[i]+ right[i];
// Count all pair of leaf nodes
// which are at most K distance apart
for(int l = 1;l < left.size(); l++)
{
for(int r = 0;r < right.size(); r++)
{
if (l + r <= distance)
{
result += left[l] * right[r];
}
}
}
// Return res to parent node
return res;
}
// Driver Code
int main()
{
/*
1
/ \
2 3
/
4
*/
Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
// Given distance K
int K = 3;
// Function call
dfs(root, K);
cout << result;
}
// This code is contributed by mohit kumar 29 C#
// Java implementation of the
// above approach
// Structure of a Node
class Node {
int data;
Node left, right;
// Constructor of the class
public Node(int item)
{
data = item;
left = right = null;
}
}
public class GFG {
Node root;
// Stores the count of required pairs
static int result;
// Function to perform dfs to find pair of
// leaf nodes at most K distance apart
static int[] dfs(Node root, int distance)
{
// Return empty array if node is null
if (root == null)
return new int[distance + 1];
// If node is a leaf node and return res
if (root.left == null
&& root.right == null) {
int res[] = new int[distance + 1];
res[1]++;
return res;
}
// Traverse to the left
int[] left = dfs(root.left,
distance);
// Traverse to the right
int[] right = dfs(root.right,
distance);
int res[] = new int[distance + 1];
// Update the distance between left
// and right leaf node
for (int i = res.length - 2;
i >= 1; i--) {
res[i + 1] = left[i]
+ right[i];
}
// Count all pair of leaf nodes
// which are at most K distance apart
for (int l = 1;
l < left.length; l++) {
for (int r = 0;
r < right.length; r++) {
if (l + r <= distance) {
result += left[l]
* right[r];
}
}
}
// Return res to parent node
return res;
}
// Driver Code
public static void main(String[] args)
{
GFG tree = new GFG();
/*
1
/ \
2 3
/
4
*/
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
result = 0;
// Given distance K
int K = 3;
// Function Call
dfs(tree.root, K);
System.out.println(result);
}
}输出：
# Python3 implementation of the
# above approach
# Structure of a Node
class newNode:
def __init__(self, item):
self.data = item
self.left = None
self.right = None
# Stores the count of required pairs
result = 0
# Function to perform dfs to find pair of
# leaf nodes at most K distance apart
def dfs(root, distance):
global result
# Return empty array if node is None
if (root == None):
res = [0 for i in range(distance + 1)]
return res
# If node is a leaf node and return res
if (root.left == None and root.right == None):
res = [0 for i in range(distance + 1)]
res[1] += 1
return res
# Traverse to the left
left = dfs(root.left, distance)
# Traverse to the right
right = dfs(root.right, distance)
res = [0 for i in range(distance + 1)]
# Update the distance between left
# and right leaf node
i = len(res) - 2
while(i >= 1):
res[i + 1] = left[i] + right[i]
i -= 1
# Count all pair of leaf nodes
# which are at most K distance apart
for l in range(1, len(left)):
for r in range(len(right)):
if (l + r <= distance):
result += left[l] * right[r]
# Return res to parent node
return res
# Driver Code
if __name__ == '__main__':
'''
1
/ \
2 3
/
4
'''
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(4)
# Given distance K
K = 3
# Function call
dfs(root, K)
print(result)
# This code is contributed by ipg2016107时间复杂度： O(N + E)，其中 N 是二叉树中的节点数， E 是边数。辅助空间： O(H)，其中 H 是二叉树的高度。
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live