Leetcode 222. Count Complete Tree Nodes

题意：

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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思路： 这道题刚开始写了个dfs遍历，线性的复杂度却超时了。 因此这道题复杂度应该是log级别的。

二叉树很容易想到二分， 很多二分策略都是log级别的

首先得到这个树的深度n，那么就能得到这个树的节点数目的区间为 [2^(n-1), 2^n-1]

有了这个区间，然后在这个区间内二分，即判断一个答案是否合理。判断的策略是从树顶往下查找（默认给这棵树编了号， 从1开始）

AC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:
    bool ok(TreeNode * root, int target, int left, int right)
    {
        if (root == NULL) return false;
        if(left==right && left == target) return true;
        if(left<right)
        {
            int mid = (left+right)/2;
            if(target <= mid)
            {
                return ok(root->left, target, left, mid);
            }
            else return ok(root->right, target, mid+1, right);
        }
        return false;
    }
    int countNodes(TreeNode* root) {
        if (root == NULL) return 0;
        TreeNode * ptr = root;
        int n = 0;
        while(ptr) 
        {
            n++;
            ptr = ptr->left;
        }
        int left_limit = pow(2, n-1) + 0.2;
        int right_limit = pow(2, n) - 1 + 0.2;
        int left = left_limit;
        int right = right_limit;
        int ans = 0;
        
        while(left<=right)
        {
            int mid = (left+right)/2;
            if( ok(root, mid, left_limit, right_limit) )
            {
                ans = max(ans, mid);
                left = mid+1;
            }
            else right = mid-1;
        }
        return ans;
    }
};