#Leetcode# 222. Count Complete Tree Nodes

https://leetcode.com/problems/count-complete-tree-nodes/

Given a complete binary tree, count the number of nodes.

Note:

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.

Example:

Input: 
    1
   / 
  2   3
 /   /
4  5 6

Output: 6

代码：

/**
 * 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) {
        int hl = 0, hr = 0;
        TreeNode *L = root;
        TreeNode *R = root;
        while(L) {
            hl ++;
            L = L -> left;
        }
        while(R) {
            hr ++;
            R = R -> right;
        }
        if(hl == hr)
            return pow(2, hl) - 1;
        return countNodes(root -> left) + countNodes(root -> right) + 1;
    }
};

　　之前用先序遍历把每一个节点存在数组中最后 $return$ 数组大小但是超时 

如果一棵树是完美二叉树的话那一定是完全二叉树，但是如果是完全二叉树的话不一定是完美二叉树 一棵完美二叉树的节点个数是 $pow(2, h) - 1$ （$h$ 是树的高度） 如果这棵树不是完美二叉树的话就返回左子树的节点个数加上右子树的节点个数加一（加上根节点） 所以就还是递归 写了这么多题目还是不很会用递归 菜哭惹