算法题（检查二叉树中的两个节点是否是表亲 |S2）

本文概述

C ++
Java
Python3
C#

给定一棵二叉树, 并且两个节点分别说" a"和" b", 请确定两个给定的节点是否互为表亲。
如果两个节点处于同一级别并且具有不同的父级, 则它们互为表亲。
例子:

6 /\ 35 / \/ \ 78 13 Say two node be 7 and 1, result is TRUE. Say two nodes are 3 and 5, result is FALSE. Say two nodes are 7 and 5, result is FALSE.

推荐：请尝试以下方法{IDE}首先, 在继续解决方案之前。解决方案套装1已经讨论了通过执行二叉树遍历来发现给定节点是否为表亲。通过执行级别顺序遍历可以解决该问题。想法是使用队列执行级别顺序遍历, 其中每个队列元素是一对节点和该节点的父节点。对于按级别顺序遍历访问的每个节点, 请检查该节点是第一给定节点还是第二给定节点。如果找到任何节点, 则存储该节点的父节点。在执行级别顺序遍历时, 一次遍历一个级别。如果在给定级别找到两个节点, 则将比较它们的父值以检查它们是否为同级。如果在给定级别中找到一个节点, 而未找到另一个节点, 则给定节点不是表亲。
下面是上述方法的实现：
C ++

// CPP program to check if two Nodes in // a binary tree are cousins // using level-order traversals #include < bits/stdc++.h> using namespace std; // A Binary Tree Node struct Node { int data; struct Node *left, *right; }; // A utility function to create a new // Binary Tree Node struct Node* newNode( int item) { struct Node* temp = ( struct Node*) malloc ( sizeof ( struct Node)); temp-> data = https://www.lsbin.com/item; temp-> left = temp-> right = NULL; return temp; }// Returns true if a and b are cousins, // otherwise false. bool isCousin(Node* root, Node* a, Node* b) { if (root == NULL) return false ; // To store parent of node a. Node* parA = NULL; // To store parent of node b. Node* parB = NULL; // queue to perform level order // traversal. Each element of // queue is a pair of node and // its parent. queue< pair< Node*, Node*> > q; // Dummy node to act like parent // of root node. Node* tmp = newNode(-1); // To store front element of queue. pair< Node*, Node*> ele; // Push root to queue. q.push(make_pair(root, tmp)); int levSize; while (!q.empty()) {// find number of elements in // current level. levSize = q.size(); while (levSize) {ele = q.front(); q.pop(); // check if current node is node a // or node b or not. if (ele.first-> data == a-> data) { parA = ele.second; }if (ele.first-> data == b-> data) { parB = ele.second; }// push children of current node // to queue. if (ele.first-> left) { q.push(make_pair(ele.first-> left, ele.first)); }if (ele.first-> right) { q.push(make_pair(ele.first-> right, ele.first)); }levSize--; // If both nodes are found in // current level then no need // to traverse current level further. if (parA & & parB) break ; }// Check if both nodes are siblings // or not. if (parA & & parB) { return parA != parB; }// If one node is found in current level // and another is not found, then // both nodes are not cousins. if ((parA & & !parB) || (parB & & !parA)) { return false ; } }return false ; } // Driver Code int main() { /* 1 / 23 / / / 45 67 / / 15 8 */struct Node* root = newNode(1); root-> left = newNode(2); root-> right = newNode(3); root-> left-> left = newNode(4); root-> left-> right = newNode(5); root-> left-> right-> right = newNode(15); root-> right-> left = newNode(6); root-> right-> right = newNode(7); root-> right-> left-> right = newNode(8); struct Node *Node1, *Node2; Node1 = root-> left-> left; Node2 = root-> right-> right; isCousin(root, Node1, Node2) ? puts ("Yes" ) : puts ( "No" ); return 0; }

Java

// Java program to check if two Nodes in // a binary tree are cousins // using level-order traversals import java.util.*; import javafx.util.Pair; // User defined node class class Node { int data; Node left, right; // Constructor to create a new tree node Node( int item) { data = https://www.lsbin.com/item; left = right = null ; } }class BinaryTree { Node root; // Returns true if a and b are cousins, // otherwise false. boolean isCousin(Node node, Node a, Node b) { if (node == null ) return false ; // To store parent of node a. Node parA = null ; // To store parent of node b. Node parB = null ; // queue to perform level order // traversal. Each element of // queue is a pair of node and // its parent. Queue< Pair < Node, Node> > q = new LinkedList< > (); // Dummy node to act like parent // of root node. Node tmp = new Node(- 1 ); // To store front element of queue. Pair< Node, Node> ele; // Push root to queue. q.add( new Pair < Node, Node> (node, tmp)); int levelSize; while (!q.isEmpty()) {// find number of elements in // current level. levelSize = q.size(); while (levelSize != 0 ) { ele = q.peek(); q.remove(); // check if current node is node a // or node b or not. if (ele.getKey().data == a.data) parA = ele.getValue(); if (ele.getKey().data == b.data) parB = ele.getValue(); // push children of current node // to queue. if (ele.getKey().left != null ) q.add( new Pair< Node, Node> (ele.getKey().left, ele.getKey())); if (ele.getKey().right != null ) q.add( new Pair< Node, Node> (ele.getKey().right, ele.getKey())); levelSize--; // If both nodes are found in // current level then no need // to traverse current level further. if (parA != null & & parB != null ) break ; }// Check if both nodes are siblings // or not. if (parA != null & & parB != null ) return parA != parB; // If one node is found in current level // and another is not found, then // both nodes are not cousins. if ((parA!= null & & parB== null ) || (parB!= null & & parA== null )) return false ; }return false ; }// Driver code public static void main(String args[]) { 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.left.right.right = new Node( 15 ); tree.root.right.left = new Node( 6 ); tree.root.right.right = new Node( 7 ); tree.root.right.left.right = new Node( 8 ); Node Node1, Node2; Node1 = tree.root.left.right.right; Node2 = tree.root.right.left.right; if (tree.isCousin(tree.root, Node1, Node2)) System.out.println("Yes" ); else System.out.println( "No" ); } }// This code is contributed by shubham96301

Python3

# Python3 program to check if two # Nodes in a binary tree are cousins # using level-order traversals # A Binary Tree Node class Node:def __init__( self , item): self .data = item self .left = None self .right = None# Returns True if a and b # are cousins, otherwise False. def isCousin(root, a, b): if root = = None : return False # To store parent of node a. parA = None # To store parent of node b. parB = None # queue to perform level order # traversal. Each element of queue # is a pair of node and its parent. q = [] # Dummy node to act like # parent of root node. tmp = Node( - 1 ) # Push root to queue. q.append((root, tmp)) while len (q) > 0 :# find number of elements in # current level. levSize = len (q) while levSize:ele = q.pop( 0 ) # check if current node is # node a or node b or not. if ele[ 0 ].data = = a.data: parA = ele[ 1 ] if ele[ 0 ].data = = b.data: parB = ele[ 1 ] # push children of # current node to queue. if ele[ 0 ].left: q.append((ele[ 0 ].left, ele[ 0 ])) if ele[ 0 ].right: q.append((ele[ 0 ].right, ele[ 0 ])) levSize - = 1# If both nodes are found in # current level then no need # to traverse current level further. if parA and parB: break # Check if both nodes # are siblings or not. if parA and parB: return parA ! = parB # If one node is found in current level # and another is not found, then # both nodes are not cousins. if (parA and not parB) or (parB and not parA): return False return False # Driver Code if __name__ = = '__main__' :root = Node( 1 ) root.left = Node( 2 ) root.right = Node( 3 ) root.left.left = Node( 4 ) root.left.right = Node( 5 ) root.left.right.right = Node( 15 ) root.right.left = Node( 6 ) root.right.right = Node( 7 ) root.right.left.right = Node( 8 ) Node1 = root.left.left Node2 = root.right.right if isCousin(root, Node1, Node2): print ( 'Yes' ) else : print ( 'No' )# This code is contributed by Rituraj Jain

// C# program to check if two Nodes in // a binary tree are cousins // using level-order traversals using System; using System.Collections.Generic; // User defined node class public class Node { public int data; public Node left, right; // Constructor to create a new tree node public Node( int item) { data = https://www.lsbin.com/item; left = right = null ; } } // User defined pair class public class Pair {public Node first, second; // Constructor to create a new tree node public Pair(Node first, Node second) { this .first = first; this .second = second; } }class BinaryTree { Node root; // Returns true if a and b are cousins, // otherwise false. Boolean isCousin(Node node, Node a, Node b) { if (node == null ) return false ; // To store parent of node a. Node parA = null ; // To store parent of node b. Node parB = null ; // queue to perform level order // traversal. Each element of // queue is a pair of node and // its parent. Queue< Pair > q = new Queue< Pair > (); // Dummy node to act like parent // of root node. Node tmp = new Node(-1); // To store front element of queue. Pair ele; // Push root to queue. q.Enqueue( new Pair (node, tmp)); int levelSize; while (q.Count> 0) {// find number of elements in // current level. levelSize = q.Count; while (levelSize != 0) { ele = q.Peek(); q.Dequeue(); // check if current node is node a // or node b or not. if (ele.first.data == a.data) parA = ele.second; if (ele.first.data == b.data) parB = ele.second; // push children of current node // to queue. if (ele.first.left != null ) q.Enqueue( new Pair(ele.first.left, ele.first)); if (ele.first.right != null ) q.Enqueue( new Pair(ele.first.right, ele.first)); levelSize--; // If both nodes are found in // current level then no need // to traverse current level further. if (parA != null & & parB != null ) break ; }// Check if both nodes are siblings // or not. if (parA != null & & parB != null ) return parA != parB; // If one node is found in current level // and another is not found, then // both nodes are not cousins. if ((parA != null & & parB == null ) || (parB != null & & parA == null )) return false ; }return false ; }// Driver code public static void Main(String []args) { 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.left.right.right = new Node(15); tree.root.right.left = new Node(6); tree.root.right.right = new Node(7); tree.root.right.left.right = new Node(8); Node Node1, Node2; Node1 = tree.root.left.right.right; Node2 = tree.root.right.left.right; if (tree.isCousin(tree.root, Node1, Node2)) Console.WriteLine("Yes" ); else Console.WriteLine( "No" ); } }// This code is contributed by Arnab Kundu

输出如下：