Determine whether a tree is balanced

Determine whether a tree is balanced.
Balance is defined as no two nodes has greater than one height.



Code:

use queue.size() to remember how many nodes at each layer (height)

// Iterative method to find height of Bianry Tree

int treeHeight(node *root)

{

    // Base Case

    if (root == NULL)

        return 0;

 

    // Create an empty queue for level order tarversal

    queue q;

 

    // Enqueue Root and initialize height

    q.push(root);

    int height = 0;

 

    while (1)

    {

        // nodeCount (queue size) indicates number of nodes

        // at current lelvel.

        int nodeCount = q.size();

        if (nodeCount == 0)

            return height;

 

        height++;

 

        // Dequeue all nodes of current level and Enqueue all

        // nodes of next level

        while (nodeCount > 0)

        {

            node *node = q.front();

            q.pop();

            if (node->left != NULL)

                q.push(node->left);

            if (node->right != NULL)

                q.push(node->right);

            nodeCount--;

        }

    }

}

Following method is to determine whether it is balanced


int checkheight(TreeNode* root)
{
    if (!root) return 0;
    int leftheight = checkheight(root->left);
    if (leftheight==-1) return -1;

    int rightheight = checkheight(root->right);
    if (rightheight==-1) return -1;

    int heightDiff = leftheight-rightheight;

    if ( abs(heightDiff) >1 )
        return -1;
    else
        return max(leftheight, rightheight) +1;
}

// check whether its is balanced tree
bool balancecheck(TreeNode* root)
{
    if(checkheight(root) == -1 )
        return false;
    else
        return true;
}