[LeetCode-JAVA] Count Complete Tree Nodes

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 2hnodes inclusive at the last level h.

题意：求一个完全二叉树的节点个数。

思路：找到最后一层的最后一个节点，可以判断左右节点最最左边的层数是否相同，如果相同，则左子树为满二叉树，若不同则右子树为满二叉树。

其中在求解的过程中，需要用到幂次的运算，如果用Math.pow会超时，可以考虑有位操作，但是要考虑2的0次幂的特殊情况。

代码：

public class Solution {
    public int countNodes(TreeNode root) {
        if(root == null)
            return 0;
        int left = countLevel(root.left);
        int right = countLevel(root.right);
        
        int leftpow = 2<<(left-1);
        int rightpow = 2<<(right-1);
        
        if(left == 0)   //0次幂，<<不出来
            leftpow = 1;
        if(right == 0)
            rightpow = 1;
        
        if(left == right){
            return leftpow + countNodes(root.right);
        }else
            return rightpow + countNodes(root.left);
    }
    
    public int countLevel(TreeNode root) {
        if(root == null)
            return 0;
        int count = 0;
        while(root != null) {
            count++;
            root = root.left;
        }
        
        return count;
    }
}

 

后来想了一下，有个小技巧，因为要用到2的0次幂，可以用1的1次幂来表示，因此可以将位操作换成( 1<< ) 可以大大精简代码。

代码：

public class Solution {
    public int countNodes(TreeNode root) {
        if(root == null)
            return 0;
        int left = countLevel(root.left); 
        int right = countLevel(root.right);
        
        if(left == right){
            return (1<<left) + countNodes(root.right);
        }else
            return (1<<right) + countNodes(root.left);
    }
    
    public int countLevel(TreeNode root) {
        if(root == null)
            return 0;
        int count = 0;
        while(root != null) {
            count++;
            root = root.left;
        }
        
        return count;
    }
}