Check If a Tree is Height Balanced 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

 Live Demo

#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