平衡（）二叉树

Question

I need to write a method that determines whether a binary tree is balanced. So first I'm going to have to determine the height of each side of the tree. But I'm having trouble understanding how I am going to count the max length of a subtree, without counting all the nodes in the subtree. This question is very difficult to ask for me, so you guys can understand.

// primary method
public int Height()
{
    int h = height( root );
}

// recursive method
private int height( Node x )
{
    if( x == null ) return 0;
    count++;                   
    height( x.left );
    height( x.right );
    return count;           
}

Here is my code for calculating the max height of the tree. But i dont know how to determine just the left or right sides' height, and this method seems to count the amount of nodes in the tree itself.

Answer 1

This method will not even compile, because of the return; statement - you need to return an int.

So, if(x == null) , what should you return? What is the height of an empty tree?

Once you have that figured out, imagine your root has only a left-subtree ( root.right == null ), and you know the height of the left-subtree. What should the height of the overall tree be?

What if it has only a right subtree?

Now, what if it has both?

Note that you don't need a global counting variable for any of this.

Answer 2

The height is 1 + max(left.height(), right.height()).

You should return a value instead of setting a variable (count, in your case), otherwise you'll go mad.

Answer 3

The height of a node is one more than the height of its tallest sub-node (hint: use Math.max ).

Answer 4

Adding to the hints I'll also point out that you really want to use recursion.