二叉树中从一个节点到达另一个节点的圈数
原文:https://www . geesforgeks . org/number-turns-reach-one-node-二叉树/
给定一棵二叉树和两个节点。任务是计算从二叉树的一个节点到达另一个节点所需的圈数。 示例:
Input: Below Binary Tree and two nodes
5 & 6
1
/ \
2 3
/ \ / \
4 5 6 7
/ / \
8 9 10
Output: Number of Turns needed to reach
from 5 to 6: 3
Input: For above tree if two nodes are 1 & 4
Output: Straight line : 0 turn
基于二叉树中最低共同祖先的思想 我们必须遵循这个步骤。 1…找到给定两个节点的生命周期评价 2…给定节点出现在左侧或右侧或等于生命周期评价。 …。根据上述条件，我们的程序属于两种情况之一:
Case 1:
If none of the nodes is equal to
LCA, we get these nodes either on
the left side or right side.
We call two functions for each node.
....a) if (CountTurn(LCA->right, first,
false, &Count)
|| CountTurn(LCA->left, first,
true, &Count)) ;
....b) Same for second node.
....Here Count is used to store number of
turns needed to reach the target node.
Case 2:
If one of the nodes is equal to LCA_Node,
then we count only the number of turns needed
to reached the second node.
If LCA == (Either first or second)
....a) if (countTurn(LCA->right, second/first,
false, &Count)
|| countTurn(LCA->left, second/first,
true, &Count)) ;
3……country turn 函数 的工作//如果我们移动 //左子树，我们通过 turn true 如果我们移动 //右子树 ，我们通过 false
CountTurn(LCA, Target_node, count, Turn)
// if found the key value in tree
if (root->key == key)
return true;
case 1:
If Turn is true that means we are
in left_subtree
If we going left_subtree then there
is no need to increment count
else
Increment count and set turn as false
case 2:
if Turn is false that means we are in
right_subtree
if we going right_subtree then there is
no need to increment count else
increment count and set turn as true.
// if key is not found.
return false;
以下是上述想法的实现。
C++
// C++ Program to count number of turns
// in a Binary Tree.
#include <bits/stdc++.h>
using namespace std;
// A Binary Tree Node
struct Node {
struct Node* left, *right;
int key;
};
// Utility function to create a new
// tree Node
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return temp;
}
// Utility function to find the LCA of
// two given values n1 and n2.
struct Node* findLCA(struct Node* root,
int n1, int n2)
{
// Base case
if (root == NULL)
return NULL;
// If either n1 or n2 matches with
// root's key, report the presence by
// returning root (Note that if a key
// is ancestor of other, then the
// ancestor key becomes LCA
if (root->key == n1 || root->key == n2)
return root;
// Look for keys in left and right subtrees
Node* left_lca = findLCA(root->left, n1, n2);
Node* right_lca = findLCA(root->right, n1, n2);
// If both of the above calls return
// Non-NULL, then one key is present
// in once subtree and other is present
// in other, So this node is the LCA
if (left_lca && right_lca)
return root;
// Otherwise check if left subtree or right
// subtree is LCA
return (left_lca != NULL) ? left_lca :
right_lca;
}
// function count number of turn need to reach
// given node from it's LCA we have two way to
bool CountTurn(Node* root, int key, bool turn,
int* count)
{
if (root == NULL)
return false;
// if found the key value in tree
if (root->key == key)
return true;
// Case 1:
if (turn == true) {
if (CountTurn(root->left, key, turn, count))
return true;
if (CountTurn(root->right, key, !turn, count)) {
*count += 1;
return true;
}
}
else // Case 2:
{
if (CountTurn(root->right, key, turn, count))
return true;
if (CountTurn(root->left, key, !turn, count)) {
*count += 1;
return true;
}
}
return false;
}
// Function to find nodes common to given two nodes
int NumberOFTurn(struct Node* root, int first,
int second)
{
struct Node* LCA = findLCA(root, first, second);
// there is no path between these two node
if (LCA == NULL)
return -1;
int Count = 0;
// case 1:
if (LCA->key != first && LCA->key != second) {
// count number of turns needs to reached
// the second node from LCA
if (CountTurn(LCA->right, second, false,
&Count)
|| CountTurn(LCA->left, second, true,
&Count))
;
// count number of turns needs to reached
// the first node from LCA
if (CountTurn(LCA->left, first, true,
&Count)
|| CountTurn(LCA->right, first, false,
&Count))
;
return Count + 1;
}
// case 2:
if (LCA->key == first) {
// count number of turns needs to reached
// the second node from LCA
CountTurn(LCA->right, second, false, &Count);
CountTurn(LCA->left, second, true, &Count);
return Count;
} else {
// count number of turns needs to reached
// the first node from LCA1
CountTurn(LCA->right, first, false, &Count);
CountTurn(LCA->left, first, true, &Count);
return Count;
}
}
// Driver program to test above functions
int main()
{
// Let us create binary tree given in the above
// example
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->left->left->left = newNode(8);
root->right->left->left = newNode(9);
root->right->left->right = newNode(10);
int turn = 0;
if ((turn = NumberOFTurn(root, 5, 10)))
cout << turn << endl;
else
cout << "Not Possible" << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
//A Java Program to count number of turns
//in a Binary Tree.
public class Turns_to_reach_another_node {
// making Count global such that it can get
// modified by different methods
static int Count;
// A Binary Tree Node
static class Node {
Node left, right;
int key;
// Constructor
Node(int key) {
this.key = key;
left = null;
right = null;
}
}
// Utility function to find the LCA of
// two given values n1 and n2.
static Node findLCA(Node root, int n1, int n2) {
// Base case
if (root == null)
return null;
// If either n1 or n2 matches with
// root's key, report the presence by
// returning root (Note that if a key
// is ancestor of other, then the
// ancestor key becomes LCA
if (root.key == n1 || root.key == n2)
return root;
// Look for keys in left and right subtrees
Node left_lca = findLCA(root.left, n1, n2);
Node right_lca = findLCA(root.right, n1, n2);
// If both of the above calls return
// Non-NULL, then one key is present
// in once subtree and other is present
// in other, So this node is the LCA
if (left_lca != null && right_lca != null)
return root;
// Otherwise check if left subtree or right
// subtree is LCA
return (left_lca != null) ? left_lca : right_lca;
}
// function count number of turn need to reach
// given node from it's LCA we have two way to
static boolean CountTurn(Node root, int key, boolean turn) {
if (root == null)
return false;
// if found the key value in tree
if (root.key == key)
return true;
// Case 1:
if (turn == true) {
if (CountTurn(root.left, key, turn))
return true;
if (CountTurn(root.right, key, !turn)) {
Count += 1;
return true;
}
} else // Case 2:
{
if (CountTurn(root.right, key, turn))
return true;
if (CountTurn(root.left, key, !turn)) {
Count += 1;
return true;
}
}
return false;
}
// Function to find nodes common to given two nodes
static int NumberOfTurn(Node root, int first, int second) {
Node LCA = findLCA(root, first, second);
// there is no path between these two node
if (LCA == null)
return -1;
Count = 0;
// case 1:
if (LCA.key != first && LCA.key != second) {
// count number of turns needs to reached
// the second node from LCA
if (CountTurn(LCA.right, second, false)
|| CountTurn(LCA.left, second, true))
;
// count number of turns needs to reached
// the first node from LCA
if (CountTurn(LCA.left, first, true)
|| CountTurn(LCA.right, first, false))
;
return Count + 1;
}
// case 2:
if (LCA.key == first) {
// count number of turns needs to reached
// the second node from LCA
CountTurn(LCA.right, second, false);
CountTurn(LCA.left, second, true);
return Count;
} else {
// count number of turns needs to reached
// the first node from LCA1
CountTurn(LCA.right, first, false);
CountTurn(LCA.left, first, true);
return Count;
}
}
// Driver program to test above functions
public static void main(String[] args) {
// Let us create binary tree given in the above
// example
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.left.left.left = new Node(8);
root.right.left.left = new Node(9);
root.right.left.right = new Node(10);
int turn = 0;
if ((turn = NumberOfTurn(root, 5, 10)) != 0)
System.out.println(turn);
else
System.out.println("Not Possible");
}
}
// This code is contributed by Sumit Ghosh
Python 3
# Python Program to count number of turns
# in a Binary Tree.
# A Binary Tree Node
class Node:
def __init__(self):
self.key = 0
self.left = None
self.right = None
# Utility function to create a new
# tree Node
def newNode(key: int) -> Node:
temp = Node()
temp.key = key
temp.left = None
temp.right = None
return temp
# Utility function to find the LCA of
# two given values n1 and n2.
def findLCA(root: Node, n1: int, n2: int) -> Node:
# Base case
if root is None:
return None
# If either n1 or n2 matches with
# root's key, report the presence by
# returning root (Note that if a key
# is ancestor of other, then the
# ancestor key becomes LCA
if root.key == n1 or root.key == n2:
return root
# Look for keys in left and right subtrees
left_lca = findLCA(root.left, n1, n2)
right_lca = findLCA(root.right, n1, n2)
# If both of the above calls return
# Non-NULL, then one key is present
# in once subtree and other is present
# in other, So this node is the LCA
if left_lca and right_lca:
return root
# Otherwise check if left subtree or right
# subtree is LCA
return (left_lca if left_lca is not None else right_lca)
# function count number of turn need to reach
# given node from it's LCA we have two way to
def countTurn(root: Node, key: int, turn: bool) -> bool:
global count
if root is None:
return False
# if found the key value in tree
if root.key == key:
return True
# Case 1:
if turn is True:
if countTurn(root.left, key, turn):
return True
if countTurn(root.right, key, not turn):
count += 1
return True
# Case 2:
else:
if countTurn(root.right, key, turn):
return True
if countTurn(root.left, key, not turn):
count += 1
return True
return False
# Function to find nodes common to given two nodes
def numberOfTurn(root: Node, first: int, second: int) -> int:
global count
LCA = findLCA(root, first, second)
# there is no path between these two node
if LCA is None:
return -1
count = 0
# case 1:
if LCA.key != first and LCA.key != second:
# count number of turns needs to reached
# the second node from LCA
if countTurn(LCA.right, second, False) or countTurn(
LCA.left, second, True):
pass
# count number of turns needs to reached
# the first node from LCA
if countTurn(LCA.left, first, True) or countTurn(
LCA.right, first, False):
pass
return count + 1
# case 2:
if LCA.key == first:
# count number of turns needs to reached
# the second node from LCA
countTurn(LCA.right, second, False)
countTurn(LCA.left, second, True)
return count
else:
# count number of turns needs to reached
# the first node from LCA1
countTurn(LCA.right, first, False)
countTurn(LCA.left, first, True)
return count
# Driver Code
if __name__ == "__main__":
count = 0
# Let us create binary tree given in the above
# example
root = 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.left.left.left = newNode(8)
root.right.left.left = newNode(9)
root.right.left.right = newNode(10)
turn = numberOfTurn(root, 5, 10)
if turn:
print(turn)
else:
print("Not possible")
# This code is contributed by
# sanjeev2552
C
using System;
//A C# Program to count number of turns
//in a Binary Tree.
public class Turns_to_reach_another_node
{
// making Count global such that it can get
// modified by different methods
public static int Count;
// A Binary Tree Node
public class Node
{
public Node left, right;
public int key;
// Constructor
public Node(int key)
{
this.key = key;
left = null;
right = null;
}
}
// Utility function to find the LCA of
// two given values n1 and n2.
public static Node findLCA(Node root, int n1, int n2)
{
// Base case
if (root == null)
{
return null;
}
// If either n1 or n2 matches with
// root's key, report the presence by
// returning root (Note that if a key
// is ancestor of other, then the
// ancestor key becomes LCA
if (root.key == n1 || root.key == n2)
{
return root;
}
// Look for keys in left and right subtrees
Node left_lca = findLCA(root.left, n1, n2);
Node right_lca = findLCA(root.right, n1, n2);
// If both of the above calls return
// Non-NULL, then one key is present
// in once subtree and other is present
// in other, So this node is the LCA
if (left_lca != null && right_lca != null)
{
return root;
}
// Otherwise check if left subtree or right
// subtree is LCA
return (left_lca != null) ? left_lca : right_lca;
}
// function count number of turn need to reach
// given node from it's LCA we have two way to
public static bool CountTurn(Node root, int key, bool turn)
{
if (root == null)
{
return false;
}
// if found the key value in tree
if (root.key == key)
{
return true;
}
// Case 1:
if (turn == true)
{
if (CountTurn(root.left, key, turn))
{
return true;
}
if (CountTurn(root.right, key, !turn))
{
Count += 1;
return true;
}
}
else // Case 2:
{
if (CountTurn(root.right, key, turn))
{
return true;
}
if (CountTurn(root.left, key, !turn))
{
Count += 1;
return true;
}
}
return false;
}
// Function to find nodes common to given two nodes
public static int NumberOfTurn(Node root, int first, int second)
{
Node LCA = findLCA(root, first, second);
// there is no path between these two node
if (LCA == null)
{
return -1;
}
Count = 0;
// case 1:
if (LCA.key != first && LCA.key != second)
{
// count number of turns needs to reached
// the second node from LCA
if (CountTurn(LCA.right, second, false)
|| CountTurn(LCA.left, second, true))
{
;
}
// count number of turns needs to reached
// the first node from LCA
if (CountTurn(LCA.left, first, true)
|| CountTurn(LCA.right, first, false))
{
;
}
return Count + 1;
}
// case 2:
if (LCA.key == first)
{
// count number of turns needs to reached
// the second node from LCA
CountTurn(LCA.right, second, false);
CountTurn(LCA.left, second, true);
return Count;
}
else
{
// count number of turns needs to reached
// the first node from LCA1
CountTurn(LCA.right, first, false);
CountTurn(LCA.left, first, true);
return Count;
}
}
// Driver program to test above functions
public static void Main(string[] args)
{
// Let us create binary tree given in the above
// example
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.left.left.left = new Node(8);
root.right.left.left = new Node(9);
root.right.left.right = new Node(10);
int turn = 0;
if ((turn = NumberOfTurn(root, 5, 10)) != 0)
{
Console.WriteLine(turn);
}
else
{
Console.WriteLine("Not Possible");
}
}
}
// This code is contributed by Shrikant13
java 描述语言
<script>
//A Javascript Program to count number of turns
//in a Binary Tree.
// making Count global such that it can get
// modified by different methods
let Count;
class Node
{
constructor(key) {
this.left = null;
this.right = null;
this.key = key;
}
}
// Utility function to find the LCA of
// two given values n1 and n2.
function findLCA(root, n1, n2) {
// Base case
if (root == null)
return null;
// If either n1 or n2 matches with
// root's key, report the presence by
// returning root (Note that if a key
// is ancestor of other, then the
// ancestor key becomes LCA
if (root.key == n1 || root.key == n2)
return root;
// Look for keys in left and right subtrees
let left_lca = findLCA(root.left, n1, n2);
let right_lca = findLCA(root.right, n1, n2);
// If both of the above calls return
// Non-NULL, then one key is present
// in once subtree and other is present
// in other, So this node is the LCA
if (left_lca != null && right_lca != null)
return root;
// Otherwise check if left subtree or right
// subtree is LCA
return (left_lca != null) ? left_lca : right_lca;
}
// function count number of turn need to reach
// given node from it's LCA we have two way to
function CountTurn(root, key, turn) {
if (root == null)
return false;
// if found the key value in tree
if (root.key == key)
return true;
// Case 1:
if (turn == true) {
if (CountTurn(root.left, key, turn))
return true;
if (CountTurn(root.right, key, !turn)) {
Count += 1;
return true;
}
} else // Case 2:
{
if (CountTurn(root.right, key, turn))
return true;
if (CountTurn(root.left, key, !turn)) {
Count += 1;
return true;
}
}
return false;
}
// Function to find nodes common to given two nodes
function NumberOfTurn(root, first, second) {
let LCA = findLCA(root, first, second);
// there is no path between these two node
if (LCA == null)
return -1;
Count = 0;
// case 1:
if (LCA.key != first && LCA.key != second) {
// count number of turns needs to reached
// the second node from LCA
if (CountTurn(LCA.right, second, false)
|| CountTurn(LCA.left, second, true))
;
// count number of turns needs to reached
// the first node from LCA
if (CountTurn(LCA.left, first, true)
|| CountTurn(LCA.right, first, false))
;
return Count + 1;
}
// case 2:
if (LCA.key == first) {
// count number of turns needs to reached
// the second node from LCA
CountTurn(LCA.right, second, false);
CountTurn(LCA.left, second, true);
return Count;
} else {
// count number of turns needs to reached
// the first node from LCA1
CountTurn(LCA.right, first, false);
CountTurn(LCA.left, first, true);
return Count;
}
}
// Let us create binary tree given in the above
// example
let 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.left.left.left = new Node(8);
root.right.left.left = new Node(9);
root.right.left.right = new Node(10);
let turn = 0;
if ((turn = NumberOfTurn(root, 5, 10)) != 0)
document.write(turn);
else
document.write("Not Possible");
// This code is contributed by suresh07.
</script>
Output
4
时间复杂度: O(n)
备选方案:
我们必须按部就班。
- 求给定两个节点的生命周期评价
- 在字符串 S1 和 S2 中存储从 LCA 开始的两个节点的路径，如果我们必须在从 LCA 开始的路径中左转到该节点，则存储‘l’，如果我们在从 LCA 开始的路径中右转，则存储“r”。
- 反转其中一个字符串并连接两个字符串。
- 计算结果字符串中不等于其相邻字符的时间字符数。
C++
// C++ Program to count number of turns
// in a Binary Tree.
#include <bits/stdc++.h>
using namespace std;
// A Binary Tree Node
struct Node {
struct Node* left, *right;
int key;
};
// Utility function to create a new
// tree Node
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return temp;
}
//Preorder traversal to store l or r in the string traversing
//from LCA to the given node key
void findPath(struct Node* root, int d,string str,string &s){
if(root==NULL){
return;
}
if(root->key==d){
s=str;
return;
}
findPath(root->left,d,str+"l",s);
findPath(root->right,d,str+"r",s);
}
// Utility function to find the LCA of
// two given values n1 and n2.
struct Node* findLCAUtil(struct Node* root, int n1, int n2,bool&v1,bool&v2){
// Base case
if (root == NULL)
return NULL;
if (root->key == n1){
v1=true;
return root;
}
if(root->key == n2){
v2=true;
return root;
}
// Look for keys in left and right subtrees
Node* left_lca = findLCAUtil(root->left, n1, n2,v1,v2);
Node* right_lca = findLCAUtil(root->right, n1, n2,v1,v2);
if (left_lca && right_lca){
return root;
}
return (left_lca != NULL) ? left_lca : right_lca;
}
bool find(Node* root, int x){
if(root==NULL){
return false;
}
if((root->key==x) || find(root->left,x) || find(root->right,x)){
return true;
}
return false;
}
//Function should return LCA of two nodes if both nodes are
//present otherwise should return NULL
Node* findLCA(struct Node* root, int first, int second){
bool v1=false;
bool v2=false;
Node* lca = findLCAUtil(root,first,second,v1,v2);
if((v1&&v2) || (v1&&find(lca,second)) || (v2&&find(lca,first))){
return lca;
}
return NULL;
}
// function should return the number of turns required to go from
//first node to second node
int NumberOFTurns(struct Node* root, int first, int second){
// base cases if root is not present or both node values are same
if(root==NULL || (first==second)){
return 0;
}
string s1 ="";
string s2 = "";
Node* lca = findLCA(root,first,second);
findPath(lca,first,"",s1);
findPath(lca,second,"",s2);
if(s1.length()==0 && s2.length()==0){
return -1;
}
reverse(s1.begin(),s1.end());
s1+=s2;
int cnt=0;
bool flag=false;
for(int i=0; i<(s1.length()-1); i++){
if(s1[i]!=s1[i+1]){
flag=true;
cnt+=1;
}
}
if(!flag){
return -1;
}
return cnt;
}
// Driver program to test above functions
int main()
{
// Let us create binary tree given in the above
// example
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->left->left->left = newNode(8);
root->right->left->left = newNode(9);
root->right->left->right = newNode(10);
int turn = 0;
if ((turn = NumberOFTurns(root, 5, 10))!=-1)
cout << turn << endl;
else
cout << "Not Possible" << endl;
return 0;
}
Output
4
时间复杂度 : O(n)
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