222. Count Complete Tree Nodes
problem description
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
algorithm thought
第一次做这个题的时候,直接没考虑这是个完全二叉树,直接写了个计算二叉树节点数量的函数,直接提交,对了就没管了。
这次做的时候,觉得肯定有更好的办法,因为完全二叉树和普通二叉树相比,多了太多信息。
这里用递归结题,我们知道,计算一个完美二叉树(忘了具体是啥名字了,就是最后一层都是NULL,其他节点都不为NULL)的节点数量很简单,如果有h层,结果就是2^h-1。对于一个完全二叉树,他左子树和右子树肯定有一个是完美二叉树,只要找到这个完美二叉树,就可以直接计算出结果,然后对另一个子树递归调用这个函数。直到节点为NULL
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) {
int h=height(root);
return h==-1?0:
height(root->right)==h-1?(1<<h)+countNodes(root->right):
(1<<(h-1))+countNodes(root->left);
}
int height(TreeNode* root){
return root==NULL?-1:height(root->left)+1; //由于保证是完全二叉树,所以计算高度直接在左边即可
}
};
algorithm analysis
这个时间复杂度是O(lgn*lgn),具体时间复杂度分析如下
T(n) = T(n/2) + c1 lgn
= T(n/4) + c1 lgn + c2 (lgn - 1)
= ...
= T(1) + c [lgn + (lgn-1) + (lgn-2) + ... + 1]
= O(lgn*lgn)
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