LeetCode Count Complete Tree Nodes

原题链接在这里：https://leetcode.com/problems/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 2h nodes inclusive at the last level h.

题解：

Complete Tree的定义是左右深度差不大于1.

递归调用函数，终止条件两个，一个是root == null, return 0. 另一个是左右高度相同说明是满树, return 2^height-1.

若是左右高度不同，递归调用求 左子树包含Node数 + 右子树包含Node数 + 1(自身).

Note:1. 用Math.pow(), 注意arguments 和 return type 都是double, 此处会TLE.

2. 用<<来完成幂运算，但注意<<的precedence比+，-还要低，需要加括号.

Time Complexity: O(h^2). worst case对于每一层都要求一遍当前点到leaf得深度.

假设只有最底层多了一个TreeNode 那么计算height就需要每层计算一遍h + (h-1) + (h-2) + ... + 1 = O(h^2).

Space: O(h), stack space.

AC  Java:

 /**
  * 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;
         }
         TreeNode p = root;
         TreeNode q = root;
         int leftDepth = 0;
         int rightDepth = 0;
         while(p!=null){
             leftDepth++;
             p = p.left;
         }
         while(q!=null){
             rightDepth++;
             q = q.right;
         }

         if(leftDepth == rightDepth){
             return (1<<leftDepth) - 1;
         }else{
             return countNodes(root.left) + 1 + countNodes(root.right);
         }
     }
 }