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：

 /**
  * 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 l = getLeft(root) + 1;
         int r = getRight(root) + 1;  

         if(l==r) {
             return (2<<(l-1)) - 1;
         } else {
             return countNodes(root.left) + countNodes(root.right) + 1;
         }
     }  

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

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