[leetcode-222]Count Complete Tree Nodes(c)-优快云博客

本文链接：https://blog.youkuaiyun.com/zdavb/article/details/47984211

本文介绍了一种高效计算完全二叉树中节点数量的方法，通过比较左右子树的深度来减少遍历次数，避免了传统暴力搜索的时间复杂度过高的问题。

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问题描述：
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.

分析：这道题如果使用暴力搜索的话，时间复杂度为O(N)，结果会TLE，然后改用一种递归的方式，每次对一个节点，求它的左节点的深度是否与右节点深度相同，如果相同，那以此为根节点的子树的个数为2^height-1，如果不同，那么就查左子树和右子树之和再加根节点。

代码如下:296ms

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int getHeight(struct TreeNode* root,bool isRight){
    int count = 0;
    while(root)
    {
        count++;
        if(isRight)
            root=root->right;
        else
            root=root->left;
    }
    return count;
}
int countNodes(struct TreeNode* root) {
    if(!root)
        return 0;
    int leftHeight = getHeight(root,false);//左子树的深度
    int rightHeight = getHeight(root,true);//右子树的深度

    if(leftHeight==rightHeight){
        return pow(2,leftHeight)-1;
    }else{
        return countNodes(root->left)+countNodes(root->right)+1;
    }
}