给定一个二叉树（每个节点最多有2个子节点），其中每个节点的值为0或1。将给定的二叉树转换为具有逻辑AND属性的树，即每个节点值都应该是其子节点之间的逻辑AND。
null
例如：
Input : The below tree doesn’t hold the logical AND property convert it to a tree that holds the property. 1 / 1 0 / / 0 1 1 1 Output : 0 / 0 1 / / 0 1 1 1
其思想是以后序方式遍历给定的二叉树。对于每个节点检查（递归），如果节点有一个子节点，那么我们不需要检查其他节点，如果节点有两个子节点，那么只需使用其子数据的逻辑AND更新节点数据。
C++
// C++ code to convert a given binary tree // to a tree that holds logical AND property. #include<bits/stdc++.h> using namespace std; // Structure of binary tree struct Node { int data; struct Node* left; struct Node* right; }; // function to create a new node struct Node* newNode( int key) { struct Node* node = new Node; node->data= key; node->left = node->right = NULL; return node; } // Convert the given tree to a tree where // each node is logical AND of its children // The main idea is to do Postorder traversal void convertTree(Node *root) { if (root == NULL) return ; /* first recur on left child */ convertTree(root->left); /* then recur on right child */ convertTree(root->right); if (root->left != NULL && root->right != NULL) root->data = (root->left->data) & (root->right->data); } void printInorder(Node* root) { if (root == NULL) return ; /* first recur on left child */ printInorder(root->left); /* then print the data of node */ printf ( "%d " , root->data); /* now recur on right child */ printInorder(root->right); } // main function int main() { /* Create following Binary Tree 1 / 1 0 / / 0 1 1 1 */ Node *root=newNode(0); root->left=newNode(1); root->right=newNode(0); root->left->left=newNode(0); root->left->right=newNode(1); root->right->left=newNode(1); root->right->right=newNode(1); printf ( " Inorder traversal before conversion " ); printInorder(root); convertTree(root); printf ( " Inorder traversal after conversion " ); printInorder(root); return 0; } |
JAVA
// Java code to convert a given binary tree // to a tree that holds logical AND property. class GfG { // Structure of binary tree static class Node { int data; Node left; Node right; } // function to create a new node static Node newNode( int key) { Node node = new Node(); node.data= key; node.left = null ; node.right = null ; return node; } // Convert the given tree to a tree where // each node is logical AND of its children // The main idea is to do Postorder traversal static void convertTree(Node root) { if (root == null ) return ; /* first recur on left child */ convertTree(root.left); /* then recur on right child */ convertTree(root.right); if (root.left != null && root.right != null ) root.data = (root.left.data) & (root.right.data); } static void printInorder(Node root) { if (root == null ) return ; /* first recur on left child */ printInorder(root.left); /* then print the data of node */ System.out.print(root.data + " " ); /* now recur on right child */ printInorder(root.right); } // main function public static void main(String[] args) { /* Create following Binary Tree 1 / 1 0 / / 0 1 1 1 */ Node root=newNode( 0 ); root.left=newNode( 1 ); root.right=newNode( 0 ); root.left.left=newNode( 0 ); root.left.right=newNode( 1 ); root.right.left=newNode( 1 ); root.right.right=newNode( 1 ); System.out.print( "Inorder traversal before conversion " ); printInorder(root); convertTree(root); System.out.println(); System.out.print( "Inorder traversal after conversion " ); printInorder(root); }} |
Python3
# Program to convert an aribitary binary tree # to a tree that holds children sum property # Helper function that allocates a new # node with the given data and None # left and right poers. class newNode: # Construct to create a new node def __init__( self , key): self .data = key self .left = None self .right = None # Convert the given tree to a tree where # each node is logical AND of its children # The main idea is to do Postorder traversal def convertTree(root) : if (root = = None ) : return """ first recur on left child """ convertTree(root.left) """ then recur on right child """ convertTree(root.right) if (root.left ! = None and root.right ! = None ): root.data = ((root.left.data) & (root.right.data)) def printInorder(root) : if (root = = None ) : return """ first recur on left child """ printInorder(root.left) """ then print the data of node """ print ( root.data, end = " " ) """ now recur on right child """ printInorder(root.right) # Driver Code if __name__ = = '__main__' : """ Create following Binary Tree 1 / 1 0 / / 0 1 1 1 """ root = newNode( 0 ) root.left = newNode( 1 ) root.right = newNode( 0 ) root.left.left = newNode( 0 ) root.left.right = newNode( 1 ) root.right.left = newNode( 1 ) root.right.right = newNode( 1 ) print ( "Inorder traversal before conversion" , end = " " ) printInorder(root) convertTree(root) print ( "Inorder traversal after conversion " , end = " " ) printInorder(root) # This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
C#
// C# code to convert a given binary tree // to a tree that holds logical AND property. using System; class GfG { // Structure of binary tree class Node { public int data; public Node left; public Node right; } // function to create a new node static Node newNode( int key) { Node node = new Node(); node.data= key; node.left = null ; node.right = null ; return node; } // Convert the given tree to a tree where // each node is logical AND of its children // The main idea is to do Postorder traversal static void convertTree(Node root) { if (root == null ) return ; /* first recur on left child */ convertTree(root.left); /* then recur on right child */ convertTree(root.right); if (root.left != null && root.right != null ) root.data = (root.left.data) & (root.right.data); } static void printInorder(Node root) { if (root == null ) return ; /* first recur on left child */ printInorder(root.left); /* then print the data of node */ Console.Write(root.data + " " ); /* now recur on right child */ printInorder(root.right); } // Driver code public static void Main() { /* Create following Binary Tree 1 / 1 0 / / 0 1 1 1 */ Node root=newNode(0); root.left=newNode(1); root.right=newNode(0); root.left.left=newNode(0); root.left.right=newNode(1); root.right.left=newNode(1); root.right.right=newNode(1); Console.Write( "Inorder traversal before conversion " ); printInorder(root); convertTree(root); Console.WriteLine(); Console.Write( "Inorder traversal after conversion " ); printInorder(root); } } /* This code is contributed by Rajput-Ji*/ |
Javascript
<script> // Javascript code to convert a given binary tree // to a tree that holds logical AND property. // Structure of binary tree class Node { constructor() { this .data = 0; this .left = null ; this .right = null ; } } // Function to create a new node function newNode(key) { var node = new Node(); node.data = key; node.left = null ; node.right = null ; return node; } // Convert the given tree to a tree where // each node is logical AND of its children // The main idea is to do Postorder traversal function convertTree(root) { if (root == null ) return ; /* First recur on left child */ convertTree(root.left); /* Then recur on right child */ convertTree(root.right); if (root.left != null && root.right != null ) root.data = (root.left.data) & (root.right.data); } function printInorder(root) { if (root == null ) return ; /* First recur on left child */ printInorder(root.left); /* then print the data of node */ document.write(root.data + " " ); /* Now recur on right child */ printInorder(root.right); } // Driver code /* Create following Binary Tree 1 / 1 0 / / 0 1 1 1 */ var root = newNode(0); root.left = newNode(1); root.right = newNode(0); root.left.left = newNode(0); root.left.right = newNode(1); root.right.left = newNode(1); root.right.right = newNode(1); document.write( "Inorder traversal before conversion " ); printInorder(root); convertTree(root); document.write( "<br>" ); document.write( "Inorder traversal after conversion " ); printInorder(root); // This code is contributed by noob2000 </script> |
输出：
Inorder traversal before conversion 0 1 1 0 1 0 1 Inorder traversal after conversion 0 0 1 0 1 1 1
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