leetCode:Count Complete Tree Nodes

计算一个二叉树的所有node节点数
刚开始直接递归遍历，后来直接超时了，因为是二叉树，可以二分的查找高度，根据高度来计算出二叉树的节点数目

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int countNodes(TreeNode root) {
        int total = 0;
        while(root != null){
            int leftLeft = level(root.left, true);
            int leftRight = level(root.left, false);
            int rightLeft = level(root.right, true);
            int rightRight = level(root.right, false);
            total += 1;
            if(leftLeft == leftRight){
                total += Math.pow(2, leftRight) - 1;
                root = root.right;
            }
            else{
                total += (int)(Math.pow(2, rightRight)) - 1;
                root = root.left;
            }
        }
        return total;
    }
    //true is left false is right
    public static int level(TreeNode root, boolean left){
        int level = 0;
        while(root != null){
            if(left) root = root.left;
            else root = root.right;
            level++;
        }
        return level;
    }
}