最近自己犯了一個愚蠢至極的錯誤，被問到如何計算一顆平衡二叉樹的高度，居然一時沒想起來，其實答案相當(dāng)簡單，用句老師們經(jīng)常愛講的話就是，這是道送分題啊，結(jié)果自己居然沒把握住...(真恨不得弄死自己)

言歸正傳，如果知道一棵平衡樹的元素數(shù)目m，則二叉樹的高度為log2m,由此可寫出代碼

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));
    }
}