[LeetCode] 222. Count Complete Tree Nodes Java

题目：

Given a complete binary tree, count the number of nodes.

Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

题意及分析：给出一颗完全二叉树，要求二叉树的节点数。完全二叉树的定义为：叶节点只能出现在最下层和次下层，并且最下面一层的结点都集中在该层最左边的若干位置的二叉树。直接遍历统计会超时。 如果从某节点一直向左的高度 = 一直向右的高度, 那么以该节点为root的子树一定是complete binary tree. 而 complete binary tree的节点数,可以用公式算出 2^h - 1. 如果高度不相等, 则递归调用 return countNode(left) + countNode(right) + 1.  复杂度为O(h^2)  

代码：

/**
 * 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) {
        if(root==null) return 0;
        int left=getLeft(root)+1;
        int right = getRight(root)+1;
        if(left==right)
            return (2<<(left-1))-1;
        else{
            return countNodes(root.left)+countNodes(root.right)+1;
        }
    }

    public int getLeft(TreeNode node){
        int count = 0;
        while(node.left!=null){
            node=node.left;
            count++;
        }
        return count;
    }

    public int getRight(TreeNode node){
        int count = 0;
        while (node.right!=null){
            node=node.right;
            count++;
        }
        return count;
    }
}