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Program to check whether a tree is height balanced or not in C++
Suppose we have a binary tree; we have to check whether its height is balanced or not. We know that for a height balanced tree, for every node in the tree, the absolute difference of the height of its left subtree and the height of its right subtree is 0 or 1.
So, if the input is like
then the output will be True
To solve this, we will follow these steps −
Define a function dfs(), this will take node,
if node is null, then −
return 0
l := 1 + dfs(left of node)
r := 1 + dfs(right of node)
if |l - r| > 1, then −
ret := false
return maximum of l and r
From the main method do the following −
ret := true
dfs(root)
return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class TreeNode {
public:
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
bool ret;
int dfs(TreeNode* node){
if(!node)
return 0;
int l = 1 + dfs(node->left);
int r = 1 + dfs(node->right);
if(abs(l - r) > 1)
ret = false;
return max(l, r);
}
bool isBalanced(TreeNode* root) {
ret = true;
dfs(root);
return ret;
}
};
main(){
Solution ob;
TreeNode *root = new TreeNode(25);
root->left = new TreeNode(19);
root->right = new TreeNode(4);
root->left->left = new TreeNode(9);
root->left->right = new TreeNode(7);
cout << (ob.isBalanced(root));
}Input
TreeNode *root = new TreeNode(25); root->left = new TreeNode(19); root->right = new TreeNode(4); root->left->left = new TreeNode(9); root->left->right = new TreeNode(7);
Output
1
Updated on: 08-Oct-2020
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