给定一棵二叉树，任务是找到一棵二叉树中两个键之间的距离，没有给出父指针。两个节点之间的距离是要遍历才能到达另一个节点的最小边数。
这个问题已经在前面的文章中讨论过了，但是它使用了二叉树的三个遍历，一个遍历用于查找两个节点(let A和B)的最低公共祖先(LCA)，然后两次遍历以查找LCA与A和LCA之间的距离B具有O(n)时间复杂度。在这篇文章中，将讨论一种方法，该方法需要O(log(n))时间来找到两个节点的LCA。
%20method_0.jpg)
两个节点之间的距离可以根据最低的共同祖先获得。以下是公式。
Dist(n1, n2) = Dist(root, n1) + Dist(root, n2) - 2*Dist(root, lca)
'n1' and 'n2' are the two given keys
'root' is root of given Binary Tree.
'lca' is lowest common ancestor of n1 and n2
Dist(n1, n2) is the distance between n1 and n2.
上面的公式也可以写成：
Dist(n1, n2) = Level[n1] + Level[n2] - 2*Level[lca]
此问题可以分解为：
- 查找每个节点的级别
- 寻找二叉树的欧拉之旅
- LCA的构建段树，
这些步骤说明如下：
- Find the levels of each node by applying level order traversal.
- Find the LCA of two nodes in binary tree in O(logn) by Storing Euler tour of Binary tree in array and computing two other arrays with the help of levels of each node and Euler tour.
These steps are shown below:
(I) First, find Euler Tour of binary tree.
%20method_1.jpg)
Euler tour of binary tree in example
- (II) Then, store levels of each node in Euler array in a different array.
%20method_2.jpg)
- (III) Then, store First occurrences of all nodes of binary tree in Euler array. H stores the indices of nodes from Euler array, so that range of query for finding minimum can be minimized and their by further optimizing the query time.
%20method_3.jpg)
- Then build segment tree on L array and take the low and high values from H array that will give us the first occurrences of say Two nodes(A and B) . Then, we query segment tree to find the minimum value say X in range (H[A] to H[B]). Then we use the index of value X as index to Euler array to get LCA, i.e. Euler[index(X)].
Let, A = 8 and B = 5.
(I) H[8] = 1 and H[5] =2
(II) Querying on Segment tree, we get min value in L array between 1 and 2 as X=0, index=7
(III) Then, LCA= Euler[7], i.e LCA = 1. - Finally, we apply distance formula discussed above to get distance between two nodes.
C++
// C++ program to find distance between
// two nodes for multiple queries
#include
#define MAX 100001
using namespace std;
/* A tree node structure */
struct Node {
int data;
struct Node* left;
struct Node* right;
};
/* Utility function to create a new Binary Tree node */
struct Node* newNode(int data)
{
struct Node* temp = new struct Node;
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// Array to store level of each node
int level[MAX];
// Utility Function to store level of all nodes
void FindLevels(struct Node* root)
{
if (!root)
return;
// queue to hold tree node with level
queue > q;
// let root node be at level 0
q.push({ root, 0 });
pair p;
// Do level Order Traversal of tree
while (!q.empty()) {
p = q.front();
q.pop();
// Node p.first is on level p.second
level[p.first->data] = p.second;
// If left child exits, put it in queue
// with current_level +1
if (p.first->left)
q.push({ p.first->left, p.second + 1 });
// If right child exists, put it in queue
// with current_level +1
if (p.first->right)
q.push({ p.first->right, p.second + 1 });
}
}
// Stores Euler Tour
int Euler[MAX];
// index in Euler array
int idx = 0;
// Find Euler Tour
void eulerTree(struct Node* root)
{
// store current node's data
Euler[++idx] = root->data;
// If left node exists
if (root->left) {
// traverse left subtree
eulerTree(root->left);
// store parent node's data
Euler[++idx] = root->data;
}
// If right node exists
if (root->right) {
// traverse right subtree
eulerTree(root->right);
// store parent node's data
Euler[++idx] = root->data;
}
}
// checks for visited nodes
int vis[MAX];
// Stores level of Euler Tour
int L[MAX];
// Stores indices of first occurrence
// of nodes in Euler tour
int H[MAX];
// Preprocessing Euler Tour for finding LCA
void preprocessEuler(int size)
{
for (int i = 1; i <= size; i++) {
L[i] = level[Euler[i]];
// If node is not visited before
if (vis[Euler[i]] == 0) {
// Add to first occurrence
H[Euler[i]] = i;
// Mark it visited
vis[Euler[i]] = 1;
}
}
}
// Stores values and positions
pair seg[4 * MAX];
// Utility function to find minimum of
// pair type values
pair min(pair a,
pair b)
{
if (a.first <= b.first)
return a;
else
return b;
}
// Utility function to build segment tree
pair buildSegTree(int low, int high, int pos)
{
if (low == high) {
seg[pos].first = L[low];
seg[pos].second = low;
return seg[pos];
}
int mid = low + (high - low) / 2;
buildSegTree(low, mid, 2 * pos);
buildSegTree(mid + 1, high, 2 * pos + 1);
seg[pos] = min(seg[2 * pos], seg[2 * pos + 1]);
}
// Utility function to find LCA
pair LCA(int qlow, int qhigh, int low,
int high, int pos)
{
if (qlow <= low && qhigh >= high)
return seg[pos];
if (qlow > high || qhigh < low)
return { INT_MAX, 0 };
int mid = low + (high - low) / 2;
return min(LCA(qlow, qhigh, low, mid, 2 * pos),
LCA(qlow, qhigh, mid + 1, high, 2 * pos + 1));
}
// Function to return distance between
// two nodes n1 and n2
int findDistance(int n1, int n2, int size)
{
// Maintain original Values
int prevn1 = n1, prevn2 = n2;
// Get First Occurrence of n1
n1 = H[n1];
// Get First Occurrence of n2
n2 = H[n2];
// Swap if low > high
if (n2 < n1)
swap(n1, n2);
// Get position of minimum value
int lca = LCA(n1, n2, 1, size, 1).second;
// Extract value out of Euler tour
lca = Euler[lca];
// return calculated distance
return level[prevn1] + level[prevn2] - 2 * level[lca];
}
void preProcessing(Node* root, int N)
{
// Build Tree
eulerTree(root);
// Store Levels
FindLevels(root);
// Find L and H array
preprocessEuler(2 * N - 1);
// Build segment Tree
buildSegTree(1, 2 * N - 1, 1);
}
/* Driver function to test above functions */
int main()
{
int N = 8; // Number of nodes
/* Constructing tree given in the above figure */
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
root->right->left->right = newNode(8);
// Function to do all preprocessing
preProcessing(root, N);
cout << "Dist(4, 5) = " <<
findDistance(4, 5, 2 * N - 1) << "\n";
cout << "Dist(4, 6) = " <<
findDistance(4, 6, 2 * N - 1) << "\n";
cout << "Dist(3, 4) = " <<
findDistance(3, 4, 2 * N - 1) << "\n";
cout << "Dist(2, 4) = " <<
findDistance(2, 4, 2 * N - 1) << "\n";
cout << "Dist(8, 5) = " <<
findDistance(8, 5, 2 * N - 1) << "\n";
return 0;
} Java
// Java program to find distance between
// two nodes for multiple queries
import java.io.*;
import java.util.*;
class GFG
{
static int MAX = 100001;
/* A tree node structure */
static class Node
{
int data;
Node left, right;
Node(int data)
{
this.data = data;
this.left = this.right = null;
}
}
static class Pair
{
T first;
V second;
Pair() {
}
Pair(T first, V second)
{
this.first = first;
this.second = second;
}
}
// Array to store level of each node
static int[] level = new int[MAX];
// Utility Function to store level of all nodes
static void findLevels(Node root)
{
if (root == null)
return;
// queue to hold tree node with level
Queue> q = new LinkedList<>();
// let root node be at level 0
q.add(new Pair(root, 0));
Pair p = new Pair();
// Do level Order Traversal of tree
while (!q.isEmpty())
{
p = q.poll();
// Node p.first is on level p.second
level[p.first.data] = p.second;
// If left child exits, put it in queue
// with current_level +1
if (p.first.left != null)
q.add(new Pair(p.first.left,
p.second + 1));
// If right child exists, put it in queue
// with current_level +1
if (p.first.right != null)
q.add(new Pair(p.first.right,
p.second + 1));
}
}
// Stores Euler Tour
static int[] Euler = new int[MAX];
// index in Euler array
static int idx = 0;
// Find Euler Tour
static void eulerTree(Node root)
{
// store current node's data
Euler[++idx] = root.data;
// If left node exists
if (root.left != null)
{
// traverse left subtree
eulerTree(root.left);
// store parent node's data
Euler[++idx] = root.data;
}
// If right node exists
if (root.right != null)
{
// traverse right subtree
eulerTree(root.right);
// store parent node's data
Euler[++idx] = root.data;
}
}
// checks for visited nodes
static int[] vis = new int[MAX];
// Stores level of Euler Tour
static int[] L = new int[MAX];
// Stores indices of first occurrence
// of nodes in Euler tour
static int[] H = new int[MAX];
// Preprocessing Euler Tour for finding LCA
static void preprocessEuler(int size)
{
for (int i = 1; i <= size; i++)
{
L[i] = level[Euler[i]];
// If node is not visited before
if (vis[Euler[i]] == 0)
{
// Add to first occurrence
H[Euler[i]] = i;
// Mark it visited
vis[Euler[i]] = 1;
}
}
}
// Stores values and positions
@SuppressWarnings("unchecked")
static Pair[] seg =
(Pair[]) new Pair[4 * MAX];
// Utility function to find minimum of
// pair type values
static Pair
min(Pair a,
Pair b)
{
if (a.first <= b.first)
return a;
return b;
}
// Utility function to build segment tree
static Pair buildSegTree(int low,
int high, int pos)
{
if (low == high)
{
seg[pos].first = L[low];
seg[pos].second = low;
return seg[pos];
}
int mid = low + (high - low) / 2;
buildSegTree(low, mid, 2 * pos);
buildSegTree(mid + 1, high, 2 * pos + 1);
seg[pos] = min(seg[2 * pos], seg[2 * pos + 1]);
return seg[pos];
}
// Utility function to find LCA
static Pair LCA(int qlow, int qhigh,
int low, int high, int pos)
{
if (qlow <= low && qhigh >= high)
return seg[pos];
if (qlow > high || qhigh < low)
return new Pair
(Integer.MAX_VALUE, 0);
int mid = low + (high - low) / 2;
return min(LCA(qlow, qhigh, low, mid, 2 * pos),
LCA(qlow, qhigh, mid + 1, high, 2 * pos + 1));
}
// Function to return distance between
// two nodes n1 and n2
static int findDistance(int n1, int n2, int size)
{
// Maintain original Values
int prevn1 = n1, prevn2 = n2;
// Get First Occurrence of n1
n1 = H[n1];
// Get First Occurrence of n2
n2 = H[n2];
// Swap if low > high
if (n2 < n1)
{
int temp = n1;
n1 = n2;
n2 = temp;
}
// Get position of minimum value
int lca = LCA(n1, n2, 1, size, 1).second;
// Extract value out of Euler tour
lca = Euler[lca];
// return calculated distance
return level[prevn1] + level[prevn2] -
2 * level[lca];
}
static void preProcessing(Node root, int N)
{
for (int i = 0; i < 4 * MAX; i++)
{
seg[i] = new Pair<>();
}
// Build Tree
eulerTree(root);
// Store Levels
findLevels(root);
// Find L and H array
preprocessEuler(2 * N - 1);
// Build segment Tree
buildSegTree(1, 2 * N - 1, 1);
}
// Driver Code
public static void main(String[] args)
{
// Number of nodes
int N = 8;
/* Constructing tree given in the above figure */
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.right.left.right = new Node(8);
// Function to do all preprocessing
preProcessing(root, N);
System.out.println("Dist(4, 5) = " +
findDistance(4, 5, 2 * N - 1));
System.out.println("Dist(4, 6) = " +
findDistance(4, 6, 2 * N - 1));
System.out.println("Dist(3, 4) = " +
findDistance(3, 4, 2 * N - 1));
System.out.println("Dist(2, 4) = " +
findDistance(2, 4, 2 * N - 1));
System.out.println("Dist(8, 5) = " +
findDistance(8, 5, 2 * N - 1));
}
}
// This code is contributed by
// sanjeev2552 Python3
# Python3 program to find distance between
# two nodes for multiple queries
from collections import deque
from sys import maxsize as INT_MAX
MAX = 100001
# A tree node structure
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Array to store level of each node
level = [0] * MAX
# Utility Function to store level of all nodes
def findLevels(root: Node):
global level
if root is None:
return
# queue to hold tree node with level
q = deque()
# let root node be at level 0
q.append((root, 0))
# Do level Order Traversal of tree
while q:
p = q[0]
q.popleft()
# Node p.first is on level p.second
level[p[0].data] = p[1]
# If left child exits, put it in queue
# with current_level +1
if p[0].left:
q.append((p[0].left, p[1] + 1))
# If right child exists, put it in queue
# with current_level +1
if p[0].right:
q.append((p[0].right, p[1] + 1))
# Stores Euler Tour
Euler = [0] * MAX
# index in Euler array
idx = 0
# Find Euler Tour
def eulerTree(root: Node):
global Euler, idx
idx += 1
# store current node's data
Euler[idx] = root.data
# If left node exists
if root.left:
# traverse left subtree
eulerTree(root.left)
idx += 1
# store parent node's data
Euler[idx] = root.data
# If right node exists
if root.right:
# traverse right subtree
eulerTree(root.right)
idx += 1
# store parent node's data
Euler[idx] = root.data
# checks for visited nodes
vis = [0] * MAX
# Stores level of Euler Tour
L = [0] * MAX
# Stores indices of the first occurrence
# of nodes in Euler tour
H = [0] * MAX
# Preprocessing Euler Tour for finding LCA
def preprocessEuler(size: int):
global L, H, vis
for i in range(1, size + 1):
L[i] = level[Euler[i]]
# If node is not visited before
if vis[Euler[i]] == 0:
# Add to first occurrence
H[Euler[i]] = i
# Mark it visited
vis[Euler[i]] = 1
# Stores values and positions
seg = [0] * (4 * MAX)
for i in range(4 * MAX):
seg[i] = [0, 0]
# Utility function to find minimum of
# pair type values
def minPair(a: list, b: list) -> list:
if a[0] <= b[0]:
return a
else:
return b
# Utility function to build segment tree
def buildSegTree(low: int, high: int,
pos: int) -> list:
if low == high:
seg[pos][0] = L[low]
seg[pos][1] = low
return seg[pos]
mid = low + (high - low) // 2
buildSegTree(low, mid, 2 * pos)
buildSegTree(mid + 1, high, 2 * pos + 1)
seg[pos] = min(seg[2 * pos], seg[2 * pos + 1])
# Utility function to find LCA
def LCA(qlow: int, qhigh: int, low: int,
high: int, pos: int) -> list:
if qlow <= low and qhigh >= high:
return seg[pos]
if qlow > high or qhigh < low:
return [INT_MAX, 0]
mid = low + (high - low) // 2
return minPair(LCA(qlow, qhigh, low, mid, 2 * pos),
LCA(qlow, qhigh, mid + 1, high, 2 * pos + 1))
# Function to return distance between
# two nodes n1 and n2
def findDistance(n1: int, n2: int, size: int) -> int:
# Maintain original Values
prevn1 = n1
prevn2 = n2
# Get First Occurrence of n1
n1 = H[n1]
# Get First Occurrence of n2
n2 = H[n2]
# Swap if low>high
if n2 < n1:
n1, n2 = n2, n1
# Get position of minimum value
lca = LCA(n1, n2, 1, size, 1)[1]
# Extract value out of Euler tour
lca = Euler[lca]
# return calculated distance
return level[prevn1] + level[prevn2] -
2 * level[lca]
def preProcessing(root: Node, N: int):
# Build Tree
eulerTree(root)
# Store Levels
findLevels(root)
# Find L and H array
preprocessEuler(2 * N - 1)
# Build sparse table
buildSegTree(1, 2 * N - 1, 1)
# Driver Code
if __name__ == "__main__":
# Number of nodes
N = 8
# Constructing tree given in the above figure
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
root.right.left.right = Node(8)
# Function to do all preprocessing
preProcessing(root, N)
print("Dist(4, 5) =",
findDistance(4, 5, 2 * N - 1))
print("Dist(4, 6) =",
findDistance(4, 6, 2 * N - 1))
print("Dist(3, 4) =",
findDistance(3, 4, 2 * N - 1))
print("Dist(2, 4) =",
findDistance(2, 4, 2 * N - 1))
print("Dist(8, 5) =",
findDistance(8, 5, 2 * N - 1))
# This code is contributed by
# sanjeev2552C#
// C# program to find distance between
// two nodes for multiple queries
using System;
using System.Collections.Generic;
class GFG
{
static int MAX = 100001;
/* A tree node structure */
public class Node
{
public int data;
public Node left, right;
public Node(int data)
{
this.data = data;
this.left = this.right = null;
}
}
class Pair
{
public T first;
public V second;
public Pair() {
}
public Pair(T first, V second)
{
this.first = first;
this.second = second;
}
}
// Array to store level of each node
static int[] level = new int[MAX];
// Utility Function to store level of all nodes
static void findLevels(Node root)
{
if (root == null)
return;
// queue to hold tree node with level
List> q =
new List>();
// let root node be at level 0
q.Add(new Pair(root, 0));
Pair p = new Pair();
// Do level Order Traversal of tree
while (q.Count != 0)
{
p = q[0];
q.RemoveAt(0);
// Node p.first is on level p.second
level[p.first.data] = p.second;
// If left child exits, put it in queue
// with current_level +1
if (p.first.left != null)
q.Add(new Pair
(p.first.left, p.second + 1));
// If right child exists, put it in queue
// with current_level +1
if (p.first.right != null)
q.Add(new Pair
(p.first.right, p.second + 1));
}
}
// Stores Euler Tour
static int[] Euler = new int[MAX];
// index in Euler array
static int idx = 0;
// Find Euler Tour
static void eulerTree(Node root)
{
// store current node's data
Euler[++idx] = root.data;
// If left node exists
if (root.left != null)
{
// traverse left subtree
eulerTree(root.left);
// store parent node's data
Euler[++idx] = root.data;
}
// If right node exists
if (root.right != null)
{
// traverse right subtree
eulerTree(root.right);
// store parent node's data
Euler[++idx] = root.data;
}
}
// checks for visited nodes
static int[] vis = new int[MAX];
// Stores level of Euler Tour
static int[] L = new int[MAX];
// Stores indices of first occurrence
// of nodes in Euler tour
static int[] H = new int[MAX];
// Preprocessing Euler Tour for finding LCA
static void preprocessEuler(int size)
{
for (int i = 1; i <= size; i++)
{
L[i] = level[Euler[i]];
// If node is not visited before
if (vis[Euler[i]] == 0)
{
// Add to first occurrence
H[Euler[i]] = i;
// Mark it visited
vis[Euler[i]] = 1;
}
}
}
// Stores values and positions
static Pair[] seg = new
Pair[4 * MAX];
// Utility function to find minimum of
// pair type values
static Pair min(Pair a,
Pair b)
{
if (a.first <= b.first)
return a;
return b;
}
// Utility function to build segment tree
static Pair buildSegTree(int low,
int high, int pos)
{
if (low == high)
{
seg[pos].first = L[low];
seg[pos].second = low;
return seg[pos];
}
int mid = low + (high - low) / 2;
buildSegTree(low, mid, 2 * pos);
buildSegTree(mid + 1, high, 2 * pos + 1);
seg[pos] = min(seg[2 * pos], seg[2 * pos + 1]);
return seg[pos];
}
// Utility function to find LCA
static Pair LCA(int qlow, int qhigh,
int low, int high, int pos)
{
if (qlow <= low && qhigh >= high)
return seg[pos];
if (qlow > high || qhigh < low)
return new Pair(int.MaxValue, 0);
int mid = low + (high - low) / 2;
return min(LCA(qlow, qhigh, low, mid, 2 * pos),
LCA(qlow, qhigh, mid + 1,
high, 2 * pos + 1));
}
// Function to return distance between
// two nodes n1 and n2
static int findDistance(int n1, int n2, int size)
{
// Maintain original Values
int prevn1 = n1, prevn2 = n2;
// Get First Occurrence of n1
n1 = H[n1];
// Get First Occurrence of n2
n2 = H[n2];
// Swap if low > high
if (n2 < n1)
{
int temp = n1;
n1 = n2;
n2 = temp;
}
// Get position of minimum value
int lca = LCA(n1, n2, 1, size, 1).second;
// Extract value out of Euler tour
lca = Euler[lca];
// return calculated distance
return level[prevn1] + level[prevn2] -
2 * level[lca];
}
static void preProcessing(Node root, int N)
{
for (int i = 0; i < 4 * MAX; i++)
{
seg[i] = new Pair();
}
// Build Tree
eulerTree(root);
// Store Levels
findLevels(root);
// Find L and H array
preprocessEuler(2 * N - 1);
// Build segment Tree
buildSegTree(1, 2 * N - 1, 1);
}
// Driver Code
public static void Main(String[] args)
{
// Number of nodes
int N = 8;
/* Constructing tree given in the above figure */
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.right.left.right = new Node(8);
// Function to do all preprocessing
preProcessing(root, N);
Console.WriteLine("Dist(4, 5) = " +
findDistance(4, 5, 2 * N - 1));
Console.WriteLine("Dist(4, 6) = " +
findDistance(4, 6, 2 * N - 1));
Console.WriteLine("Dist(3, 4) = " +
findDistance(3, 4, 2 * N - 1));
Console.WriteLine("Dist(2, 4) = " +
findDistance(2, 4, 2 * N - 1));
Console.WriteLine("Dist(8, 5) = " +
findDistance(8, 5, 2 * N - 1));
}
}
// This code is contributed by Rajput-Ji 输出：
Dist(4, 5) = 2
Dist(4, 6) = 4
Dist(3, 4) = 3
Dist(2, 4) = 1
Dist(8, 5) = 5
时间复杂度： O(Log N)
空间复杂度： O(N)
查找二叉树的两个节点之间的距离的查询– O(1)方法
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