最近自己犯了一个愚蠢至极的错误，被问到如何计算一颗平衡二叉树的高度，居然一时没想起来，其实答案相当简单，用句老师们经常爱讲的话就是，这是道送分题啊，结果自己居然没把握住...(真恨不得弄死自己)
言归正传，如果知道一棵平衡树的元素数目m，则二叉树的高度为log2m,由此可写出代码
public class BinaryTreeNode<T> {
public T element;
public BinaryTreeNode<T> left, right;
public BinaryTreeNode(T element) {
this.element = element;
this.left = null;
this.right = null;
}
public int numChildren() {
int childrenCount = 0;
if (left != null) {
childrenCount = 1 + left.numChildren();
}
if (right != null) {
childrenCount = childrenCount + 1 + right.numChildren();
}
return childrenCount;
}
public int getHeight(int nodesNum) {
return (int)(Math.log(nodesNum) / Math.log(2));
}
}
