LeetCode222 Count CompleteTree Nodes(计算全然二叉树的节点数) Java 题解

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.

解题：

假设用常规的解法一个个遍历，就是O(n)时间复杂度 。会不通过，这边就不写O(n)的代码了。由于是全然二叉树，满二叉树有一个性质是节点数等于2^h-1,h为高度，所以能够这样推断节点的左右高度是不是一样，假设是一样说明是满二叉树，就能够用刚才的公式。假设左右不相等就递归计算左右节点。

代码：

public static int countNodes(TreeNode root) {
		 if(root==null)
			 return 0;
		 else {
			int left=getLeftHeight(root);
			int right=getRightHeight(root);
			if(left==right)
				return (1<<left)-1;
			else {
				return countNodes(root.right)+countNodes(root.left)+1;
			}
		}
	 }
	 
	 public static int  getRightHeight(TreeNode root) {
		 int height=0;
		 while(root!=null)
		 {
			 height++;
			 root=root.left;
		 }
		 return height;
		
	}
	 
	 public static int  getLeftHeight(TreeNode root) {
		 int height=0;
		 while(root!=null)
		 {
			 height++;
			 root=root.right;
		 }
		 return height;
		
	}