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.

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可以证明一个完全二叉树左右子树至少有一个是满二叉树。

满二叉树的节点数是2^k-1，k是树的深度。

所以我们可以先判断该树是否为满二叉树，然后是的话直接返回结果，如果不是递归地求解子树。

这样不用遍历所有的节点。复杂度小于O(N)，比对所有点遍历复杂度要小，最好的情况是O(lgN)。

推算大概在O(lgN)~O(N)之间。

具体的分析，取左右子树只有一个是满树的最差情况。

T(N) = lg(N/2) + T(N/2)

T(1) =1 

可以推导下复杂度最差在O(lgN*lgN)。

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code:

/**
 * 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) {
        if(root==nullptr)  
            return 0;  
        TreeNode* lt = root;  
        TreeNode* rt = root;  
        int leftdepth = 0;  
        int rightdepth = 0;  
        while(lt){  
            leftdepth++;  
            lt = lt->left;  
        }  
        while(rt){  
            rightdepth++;  
            rt = rt->right;  
        }  
        if(leftdepth==rightdepth)  
            return pow(2,leftdepth)-1;  
        else  
            return countNodes(root->left)+countNodes(root->right)+1; 
    }
};