Leetcode 222: Count Complete Tree Nodes

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本文介绍了一种高效计算完全二叉树节点数量的方法，通过比较左右子树的高度来判断是否为完全二叉树，并利用数学公式快速计算节点总数，避免了递归导致的时间限制问题。

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Do only the recursion will cause TLE. Thus, at each node, count the heights of its left subtree and right subtree, if left height equals to the right height, the number of nodes of this root can be calculated in the mathematical way.

/**
 * 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)
            return 0;
        int hl=0,hr=0;
        TreeNode *l=root->left,*r=root->right;
        while(l)
        {
            l=l->left;
            hl++;
        }
        while(r)
        {
            r=r->right;
            hr++;
        }
        if(hl==hr)
            return (1<<hl+1)-1;//equals to pow(2,hl+1)-1,increase the speed
        return 1+countNodes(root->left)+countNodes(root->right);
    }
};