LeetCode 222. 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);
        }
    }
}