我们有一个简单的二叉树，我们必须打印二叉树中最大的三个元素。 例如：
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Input : 1 / 2 3 / / 4 5 4 5Output :Three largest elements are 5 4 3
方法 我们可以简单地取三个变量第一、第二、第三，分别存储第一、第二、第三大变量，并执行预序遍历，每次我们都会相应地更新元素。 这种方法只需要O（n）个时间。 算法-
1- Take 3 variables first, second, third2- Perform a preorder traversal if (root==NULL) return if root's data is larger then first update third with second second with first first with root's data else if root's data is larger then second and not equal to first update third with second second with root's data else if root's data is larger then third and not equal to first & second update third with root's data3- call preorder for root->left4- call preorder for root->right
C++
// CPP program to find largest three elements in // a binary tree. #include <bits/stdc++.h> using namespace std; struct Node { int data; struct Node *left; struct Node *right; }; /* Helper function that allocates a new Node with the given data and NULL left and right pointers. */ struct Node *newNode( int data) { struct Node *node = new Node; node->data = data; node->left = NULL; node->right = NULL; return (node); } // function to find three largest element void threelargest(Node *root, int &first, int &second, int &third) { if (root == NULL) return ; // if data is greater than first large number // update the top three list if (root->data > first) { third = second; second = first; first = root->data; } // if data is greater than second large number // and not equal to first update the bottom // two list else if (root->data > second && root->data != first) { third = second; second = root->data; } // if data is greater than third large number // and not equal to first & second update the // third highest list else if (root->data > third && root->data != first && root->data != second) third = root->data; threelargest(root->left, first, second, third); threelargest(root->right, first, second, third); } // driver function int main() { struct 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(4); root->right->right = newNode(5); int first = 0, second = 0, third = 0; threelargest(root, first, second, third); cout << "three largest elements are " << first << " " << second << " " << third; return 0; } |
JAVA
// Java program to find largest three elements // in a binary tree. import java.util.*; class GFG { static class Node { int data; Node left; Node right; }; static int first, second, third; /* Helper function that allocates a new Node with the given data and null left and right pointers. */ static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = null ; node.right = null ; return (node); } // function to find three largest element static void threelargest(Node root) { if (root == null ) return ; // if data is greater than first large number // update the top three list if (root.data > first) { third = second; second = first; first = root.data; } // if data is greater than second large number // and not equal to first update the bottom // two list else if (root.data > second && root.data != first) { third = second; second = root.data; } // if data is greater than third large number // and not equal to first & second update the // third highest list else if (root.data > third && root.data != first && root.data != second) third = root.data; threelargest(root.left); threelargest(root.right); } // driver function public static void main(String[] args) { 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( 4 ); root.right.right = newNode( 5 ); first = 0 ; second = 0 ; third = 0 ; threelargest(root); System.out.print( "three largest elements are " + first + " " + second + " " + third); } } // This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to find largest three # elements in a binary tree. # Helper function that allocates a new # Node with the given data and None # left and right pointers. class newNode: def __init__( self , data): self .data = data self .left = None self .right = None # function to find three largest element def threelargest(root, first, second, third): if (root = = None ): return # if data is greater than first large # number update the top three list if (root.data > first[ 0 ]): third[ 0 ] = second[ 0 ] second[ 0 ] = first[ 0 ] first[ 0 ] = root.data # if data is greater than second large # number and not equal to first update # the bottom two list elif (root.data > second[ 0 ] and root.data ! = first[ 0 ]): third[ 0 ] = second[ 0 ] second[ 0 ] = root.data # if data is greater than third large # number and not equal to first & second # update the third highest list elif (root.data > third[ 0 ] and root.data ! = first[ 0 ] and root.data ! = second[ 0 ]): third[ 0 ] = root.data threelargest(root.left, first, second, third) threelargest(root.right, first, second, third) # Driver Code if __name__ = = '__main__' : 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( 4 ) root.right.right = newNode( 5 ) first = [ 0 ] second = [ 0 ] third = [ 0 ] threelargest(root, first, second, third) print ( "three largest elements are" , first[ 0 ], second[ 0 ], third[ 0 ]) # This code is contributed by PranchalK |
C#
// C# program to find largest three elements // in a binary tree. using System; class GFG { public class Node { public int data; public Node left; public Node right; }; static int first, second, third; /* Helper function that allocates a new Node with the given data and null left and right pointers. */ static Node newNode( int data) { Node node = new Node(); node.data = data; node.left = null ; node.right = null ; return (node); } // function to find three largest element static void threelargest(Node root) { if (root == null ) return ; // if data is greater than first large number // update the top three list if (root.data > first) { third = second; second = first; first = root.data; } // if data is greater than second large number // and not equal to first update the bottom // two list else if (root.data > second && root.data != first) { third = second; second = root.data; } // if data is greater than third large number // and not equal to first & second update the // third highest list else if (root.data > third && root.data != first && root.data != second) third = root.data; threelargest(root.left); threelargest(root.right); } // Driver Code public static void Main(String[] args) { 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(4); root.right.right = newNode(5); first = 0; second = 0; third = 0; threelargest(root); Console.WriteLine( "three largest elements are " + first + " " + second + " " + third); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript program to find largest // three elements in a binary tree. class Node { constructor(data) { this .left = null ; this .right = null ; this .data = data; } } let first, second, third; /* Helper function that allocates a new Node with the given data and null left and right pointers. */ function newNode(data) { let node = new Node(data); return (node); } // function to find three largest element function threelargest(root) { if (root == null ) return ; // if data is greater than first large number // update the top three list if (root.data > first) { third = second; second = first; first = root.data; } // if data is greater than second large number // and not equal to first update the bottom // two list else if (root.data > second && root.data != first) { third = second; second = root.data; } // if data is greater than third large number // and not equal to first & second update the // third highest list else if (root.data > third && root.data != first && root.data != second) third = root.data; threelargest(root.left); threelargest(root.right); } let 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(4); root.right.right = newNode(5); first = 0; second = 0; third = 0; threelargest(root); document.write( "three largest elements are " + first + " " + second + " " + third); </script> |
输出：
three largest elements are 5 4 3
时间复杂性： O（N）
辅助空间： O（1）
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