222. Count Complete Tree Nodes

/*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.完全二叉树：除了最后一层，第i+1层都有2^i个节点，最后一层可能缺失从右边开始连续的几个节点等比数列求和公式：Sn = a(1-q^n)/(1-q) 2^0 + 2^1 + 2^2+ ...+2^(n-1)=2^n -1寻找最后一层从右边数第一个不为空的节点，采用二分法查找，通过本节点的层数h(同过一直计算左支点的个数)与右节点的层数h2比较来把树分成两个子树，如果h==h2-1则说明左子树为满二叉树，则转向右子树计算......*//** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int countNodes(TreeNode* root) {//Time linit exceeded        if(!root) return 0;        return countNodes(root->left)+countNodes(root->right)+1;    }    int height(TreeNode* root){        int i=0;        while(root){            root=root->left;            i++;        }        return i;    }    int countNodes(TreeNode* root) {        int h=height(root);        return h==0 ? 0 : height(root->right)==h-1 ? (1<<h-1) + countNodes(root->right) : (1<<h-2) + countNodes(root->left);    }};