Given a binary tree. The task is to check if the binary tree is sorted level-wise or not. A binary tree is level sorted if max( i-1th level) is less than min( ith level ).
Examples:
Input : 1
/ \
/ \
2 3
/ \ / \
/ \ / \
4 5 6 7
Output : Sorted
Input: 1
/
4
/ \
6 5
\
2
Output: Not sorted
Simple Solution: A simple solution is to compare minimum and maximum value of each adjacent level i and i+1. Traverse to ith and i+1th level, compare the minimum value of i+1th level with maximum value of ith level and return the result.
Time complexity: O(n2).
Efficient Solution: An efficient solution is to do level order traversal and keep track of the minimum and maximum values of current level. Use a variable prevMax to store the maximum value of the previous level. Then compare the minimum value of current level with the maximum value of the previous level, pevMax. If minimum value is greater than the prevMax, then the given tree is sorted level-wise up to current level. For next level, prevMax is the equal to maximum value of current level. So update the prevMax with maximum value of current level. Repeat this until all levels of given tree are not traversed.
Below is the implementation of above approach:
C++
// CPP program to determine whether// binary tree is level sorted or not.#include <bits/stdc++.h>using namespace std;// Structure of a tree node.struct Node { int key; Node *left, *right;};// Function to create new tree node.Node* newNode(int key){ Node* temp = new Node; temp->key = key; temp->left = temp->right = NULL; return temp;}// Function to determine if// given binary tree is level sorted// or not.int isSorted(Node* root){ // to store maximum value of previous // level. int prevMax = INT_MIN; // to store minimum value of current // level. int minval; // to store maximum value of current // level. int maxval; // to store number of nodes in current // level. int levelSize; // queue to perform level order traversal. queue<Node*> q; q.push(root); while (!q.empty()) { // find number of nodes in current // level. levelSize = q.size(); minval = INT_MAX; maxval = INT_MIN; // traverse current level and find // minimum and maximum value of // this level. while (levelSize > 0) { root = q.front(); q.pop(); levelSize--; minval = min(minval, root->key); maxval = max(maxval, root->key); if (root->left) q.push(root->left); if (root->right) q.push(root->right); } // if minimum value of this level // is not greater than maximum // value of previous level then // given tree is not level sorted. if (minval <= prevMax) return 0; // maximum value of this level is // previous maximum value for // next level. prevMax = maxval; } return 1;}// Driver programint main(){ /* 1 / 4 \ 6 / \ 8 9 / \ 12 10 */ Node* root = newNode(1); root->left = newNode(4); root->left->right = newNode(6); root->left->right->left = newNode(8); root->left->right->right = newNode(9); root->left->right->left->left = newNode(12); root->left->right->right->right = newNode(10); if (isSorted(root)) cout << "Sorted"; else cout << "Not sorted"; return 0;} |
Java
// Java program to determine whether // binary tree is level sorted or not. import java.util.*;class GfG { // Structure of a tree node. static class Node { int key; Node left, right; }// Function to create new tree node. static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.left = null; temp.right = null; return temp; } // Function to determine if // given binary tree is level sorted // or not. static int isSorted(Node root) { // to store maximum value of previous // level. int prevMax = Integer.MIN_VALUE; // to store minimum value of current // level. int minval; // to store maximum value of current // level. int maxval; // to store number of nodes in current // level. int levelSize; // queue to perform level order traversal. Queue<Node> q = new LinkedList<Node> (); q.add(root); while (!q.isEmpty()) { // find number of nodes in current // level. levelSize = q.size(); minval = Integer.MAX_VALUE; maxval = Integer.MIN_VALUE; // traverse current level and find // minimum and maximum value of // this level. while (levelSize > 0) { root = q.peek(); q.remove(); levelSize--; minval = Math.min(minval, root.key); maxval = Math.max(maxval, root.key); if (root.left != null) q.add(root.left); if (root.right != null) q.add(root.right); } // if minimum value of this level // is not greater than maximum // value of previous level then // given tree is not level sorted. if (minval <= prevMax) return 0; // maximum value of this level is // previous maximum value for // next level. prevMax = maxval; } return 1; } // Driver program public static void main(String[] args) { /* 1 / 4 \ 6 / \ 8 9 / \ 12 10 */ Node root = newNode(1); root.left = newNode(4); root.left.right = newNode(6); root.left.right.left = newNode(8); root.left.right.right = newNode(9); root.left.right.left.left = newNode(12); root.left.right.right.right = newNode(10); if (isSorted(root) == 1) System.out.println("Sorted"); else System.out.println("Not sorted"); }} |
Python3
# Python3 program to determine whether # binary tree is level sorted or not. from queue import Queue# Function to create new tree node. class newNode: def __init__(self, key): self.key = key self.left = self.right = None# Function to determine if given binary# tree is level sorted or not. def isSorted(root): # to store maximum value of previous # level. prevMax = -999999999999 # to store minimum value of current # level. minval = None # to store maximum value of current # level. maxval = None # to store number of nodes in current # level. levelSize = None # queue to perform level order traversal. q = Queue() q.put(root) while (not q.empty()): # find number of nodes in current # level. levelSize = q.qsize() minval = 999999999999 maxval = -999999999999 # traverse current level and find # minimum and maximum value of # this level. while (levelSize > 0): root = q.queue[0] q.get() levelSize -= 1 minval = min(minval, root.key) maxval = max(maxval, root.key) if (root.left): q.put(root.left) if (root.right): q.put(root.right) # if minimum value of this level # is not greater than maximum # value of previous level then # given tree is not level sorted. if (minval <= prevMax): return 0 # maximum value of this level is # previous maximum value for # next level. prevMax = maxval return 1# Driver Code if __name__ == '__main__': # # 1 # / # 4 # \ # 6 # / \ # 8 9 # / \ # 12 10 root = newNode(1) root.left = newNode(4) root.left.right = newNode(6) root.left.right.left = newNode(8) root.left.right.right = newNode(9) root.left.right.left.left = newNode(12) root.left.right.right.right = newNode(10) if (isSorted(root)): print("Sorted") else: print("Not sorted")# This code is contributed by PranchalK |
C#
// C# program to determine whether // binary tree is level sorted or not. using System;using System.Collections.Generic; class GFG { // Structure of a tree node. public class Node{ public int key; public Node left, right; }// Function to create new tree node. static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.left = null; temp.right = null; return temp; } // Function to determine if // given binary tree is level sorted // or not. static int isSorted(Node root) { // to store maximum value of previous // level. int prevMax = int.MinValue; // to store minimum value of current // level. int minval; // to store maximum value of current // level. int maxval; // to store number of nodes in current // level. int levelSize; // queue to perform level order traversal. Queue<Node> q = new Queue<Node> (); q.Enqueue(root); while (q.Count != 0) { // find number of nodes in current // level. levelSize = q.Count; minval = int.MaxValue; maxval = int.MinValue; // traverse current level and find // minimum and maximum value of // this level. while (levelSize > 0) { root = q.Peek(); q.Dequeue(); levelSize--; minval = Math.Min(minval, root.key); maxval = Math.Max(maxval, root.key); if (root.left != null) q.Enqueue(root.left); if (root.right != null) q.Enqueue(root.right); } // if minimum value of this level // is not greater than maximum // value of previous level then // given tree is not level sorted. if (minval <= prevMax) return 0; // maximum value of this level is // previous maximum value for // next level. prevMax = maxval; } return 1; } // Driver Codepublic static void Main(String[] args) { /* 1 / 4 \ 6 / \ 8 9 / \ 12 10 */ Node root = newNode(1); root.left = newNode(4); root.left.right = newNode(6); root.left.right.left = newNode(8); root.left.right.right = newNode(9); root.left.right.left.left = newNode(12); root.left.right.right.right = newNode(10); if (isSorted(root) == 1) Console.WriteLine("Sorted"); else Console.WriteLine("Not sorted"); }} // This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript program to determine whether // binary tree is level sorted or not. // Structure of a tree node. class Node{ constructor() { this.key = null; this.left = null; this.right = null; }}// Function to create new tree node. function newNode(key) { var temp = new Node(); temp.key = key; temp.left = null; temp.right = null; return temp; } // Function to determine if // given binary tree is level sorted // or not. function isSorted(root) { // to store maximum value of previous // level. var prevMax = -1000000000; // to store minimum value of current // level. var minval; // to store maximum value of current // level. var maxval; // to store number of nodes in current // level. var levelSize; // queue to perform level order traversal. var q = []; q.push(root); while (q.length != 0) { // find number of nodes in current // level. levelSize = q.length; minval = 1000000000; maxval = -1000000000; // traverse current level and find // minimum and maximum value of // this level. while (levelSize > 0) { root = q[0]; q.shift(); levelSize--; minval = Math.min(minval, root.key); maxval = Math.max(maxval, root.key); if (root.left != null) q.push(root.left); if (root.right != null) q.push(root.right); } // if minimum value of this level // is not greater than maximum // value of previous level then // given tree is not level sorted. if (minval <= prevMax) return 0; // maximum value of this level is // previous maximum value for // next level. prevMax = maxval; } return 1; } // Driver Code/* 1 / 4 \ 6 / \ 8 9 / \ 12 10 */var root = newNode(1); root.left = newNode(4); root.left.right = newNode(6); root.left.right.left = newNode(8); root.left.right.right = newNode(9); root.left.right.left.left = newNode(12); root.left.right.right.right = newNode(10); if (isSorted(root) == 1) document.write("Sorted"); else document.write("Not Sorted"); </script> |
Output:
Sorted
Time Complexity: O(n)
Auxiliary Space: O(n)
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