LeetCode OJ 222. Count Complete Tree Nodes

Total Accepted: 32628 Total Submissions: 129569 Difficulty: Medium

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.

如果用暴力递归来计算节点总数,那么会超时,时间复杂度是O(N)。由于我们求的是完全二叉树的节点数,如果用求普通二叉树的方法来解决肯定是不合适的。完全二叉树的一些特点可能成为我们解决该问题的思路:如果一个节点的左子树的深度和右子树的深度一样,那么以该节点为根节点的树的节点数为:h2-1。最好情况下的时间复杂度为O(h),其它情况的时间复杂度为O(h2)。代码如下:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int countNodes(TreeNode root) {  
12         if(root==null) return 0;  
13           
14         int l = getLeft(root) + 1;  
15         int r = getRight(root) + 1;  
16           
17         if(l==r) {  
18             return (2<<(l-1)) - 1;  
19         } else {  
20             return countNodes(root.left) + countNodes(root.right) + 1;  
21         }  
22     }  
23       
24     private int getLeft(TreeNode root) {  
25         int count = 0;  
26         while(root.left!=null) {  
27             root = root.left;  
28             ++count;  
29         }  
30         return count;  
31     }  
32       
33     private int getRight(TreeNode root) {  
34         int count = 0;  
35         while(root.right!=null) {  
36             root = root.right;  
37             ++count;  
38         }  
39         return count;  
40     }  
41 }

 

 
posted @ 2016-05-09 09:28  Black_Knight  阅读(199)  评论(0编辑  收藏  举报