打印从根(或公共祖先)开始的路径上的公共节点
原文:https://www . geesforgeks . org/print-common-nodes-path-root-common-predents/
给定一个二叉树和两个节点，任务是打印二叉树中两个给定节点的所有公共节点。
示例:
Given binary tree is :
1
/ \
2 3
/ \ / \
4 5 6 7
/ / \
8 9 10
Given nodes 9 and 7, so the common nodes are:-
1, 3
问于:亚马逊
C++
// C++ Program to find common nodes for given two nodes
#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;
}
// Utility Function to print all ancestors of LCA
bool printAncestors(struct Node* root, int target)
{
/* base cases */
if (root == NULL)
return false;
if (root->key == target) {
cout << root->key << " ";
return true;
}
/* If target is present in either left or right
subtree of this node, then print this node */
if (printAncestors(root->left, target) ||
printAncestors(root->right, target)) {
cout << root->key << " ";
return true;
}
/* Else return false */
return false;
}
// Function to find nodes common to given two nodes
bool findCommonNodes(struct Node* root, int first,
int second)
{
struct Node* LCA = findLCA(root, first, second);
if (LCA == NULL)
return false;
printAncestors(root, LCA->key);
}
// 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);
if (findCommonNodes(root, 9, 7) == false)
cout << "No Common nodes";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to find common nodes for given two nodes
import java.util.LinkedList;
// Class to represent Tree node
class Node
{
int data;
Node left, right;
public Node(int item)
{
data = item;
left = null;
right = null;
}
}
// Class to count full nodes of Tree
class BinaryTree
{
static Node root;
// 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.data == n1 || root.data == 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;
}
// Utility Function to print all ancestors of LCA
static boolean printAncestors(Node root, int target)
{
/* base cases */
if (root == null)
return false;
if (root.data == target) {
System.out.print(root.data+ " ");
return true;
}
/* If target is present in either left or right
subtree of this node, then print this node */
if (printAncestors(root.left, target) ||
printAncestors(root.right, target)) {
System.out.print(root.data+ " ");
return true;
}
/* Else return false */
return false;
}
// Function to find nodes common to given two nodes
static boolean findCommonNodes(Node root, int first,
int second)
{
Node LCA = findLCA(root, first, second);
if (LCA == null)
return false;
printAncestors(root, LCA.data);
return true;
}
// Driver program to test above functions
public static void main(String args[])
{
/*Let us create Binary Tree shown in
above example */
BinaryTree tree = new BinaryTree();
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);
tree.root.right.left = new Node(6);
tree.root.right.right = new Node(7);
tree.root.left.left.left = new Node(8);
tree.root.right.left.left = new Node(9);
tree.root.right.left.right = new Node(10);
if (findCommonNodes(root, 9, 7) == false)
System.out.println("No Common nodes");
}
}
// This code is contributed by Mr Somesh Awasthi
Python 3
# Python3 Program to find common
# nodes for given two nodes
# Utility class to create a new tree Node
class newNode:
def __init__(self, key):
self.key = key
self.left = self.right = None
# Utility function to find the LCA of
# two given values n1 and n2.
def findLCA(root, n1, n2):
# Base case
if (root == 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-None,
# 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
if (left_lca != None):
return left_lca
else:
return right_lca
# Utility Function to print all ancestors of LCA
def printAncestors(root, target):
# base cases
if (root == None):
return False
if (root.key == target):
print(root.key, end = " ")
return True
# If target is present in either left or right
# subtree of this node, then prthis node
if (printAncestors(root.left, target) or
printAncestors(root.right, target)):
print(root.key, end = " ")
return True
# Else return False
return False
# Function to find nodes common to given two nodes
def findCommonNodes(root, first, second):
LCA = findLCA(root, first, second)
if (LCA == None):
return False
printAncestors(root, LCA.key)
# Driver Code
if __name__ == '__main__':
# Let us create binary tree given
# in the above example
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)
if (findCommonNodes(root, 9, 7) == False):
print("No Common nodes")
# This code is contributed by PranchalK
C
using System;
// C# Program to find common nodes for given two nodes
// Class to represent Tree node
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = null;
right = null;
}
}
// Class to count full nodes of Tree
public class BinaryTree
{
public static Node root;
// 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.data == n1 || root.data == 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;
}
// Utility Function to print all ancestors of LCA
public static bool printAncestors(Node root, int target)
{
/* base cases */
if (root == null)
{
return false;
}
if (root.data == target)
{
Console.Write(root.data + " ");
return true;
}
/* If target is present in either left or right
subtree of this node, then print this node */
if (printAncestors(root.left, target)
|| printAncestors(root.right, target))
{
Console.Write(root.data + " ");
return true;
}
/* Else return false */
return false;
}
// Function to find nodes common to given two nodes
public static bool findCommonNodes(Node root,
int first, int second)
{
Node LCA = findLCA(root, first, second);
if (LCA == null)
{
return false;
}
printAncestors(root, LCA.data);
return true;
}
// Driver program to test above functions
public static void Main(string[] args)
{
/*Let us create Binary Tree shown in
above example */
BinaryTree tree = new BinaryTree();
BinaryTree.root = new Node(1);
BinaryTree.root.left = new Node(2);
BinaryTree.root.right = new Node(3);
BinaryTree.root.left.left = new Node(4);
BinaryTree.root.left.right = new Node(5);
BinaryTree.root.right.left = new Node(6);
BinaryTree.root.right.right = new Node(7);
BinaryTree.root.left.left.left = new Node(8);
BinaryTree.root.right.left.left = new Node(9);
BinaryTree.root.right.left.right = new Node(10);
if (findCommonNodes(root, 9, 7) == false)
{
Console.WriteLine("No Common nodes");
}
}
}
// This code is contributed by Shrikant13
java 描述语言
<script>
// Javascript program to find common
// nodes for given two nodes
class Node
{
constructor(data)
{
this.left = null;
this.right = null;
this.data = data;
}
}
let root;
// 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.data == n1 || root.data == 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;
}
// Utility Function to print all ancestors of LCA
function printAncestors(root, target)
{
// Base cases
if (root == null)
return false;
if (root.data == target)
{
document.write(root.data + " ");
return true;
}
// If target is present in either left or right
// subtree of this node, then print this node
if (printAncestors(root.left, target) ||
printAncestors(root.right, target))
{
document.write(root.data+ " ");
return true;
}
// Else return false
return false;
}
// Function to find nodes common to given two nodes
function findCommonNodes(root, first, second)
{
let LCA = findLCA(root, first, second);
if (LCA == null)
return false;
printAncestors(root, LCA.data);
return true;
}
// Driver code
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);
if (findCommonNodes(root, 9, 7) == false)
document.write("No Common nodes");
// This code is contributed by divyeshrabadiya07
</script>
输出:
3 1
本文由 萨哈布拉 供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 review-team@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
版权属于：月萌API www.moonapi.com，转载请注明出处
