leetCode(19):Count Complete Tree Nodes 分类： leetCode 2015-06-22 08:32 222人阅读 评论(0) 收藏

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 2hnodes inclusive at the last level h.

       代码一适合任何形式的二叉树，代码二和代码三只适合本题所述形式的二叉树。

代码一：最简单，但时间超限

int countNodes(TreeNode* root)
{
	if(NULL==root)
		return 0;
	int count(0);
	count=1+countNodes(root->left)+countNodes(root->right);
	return count;
}

代码二：采用层序遍历的方式，统计层数和最后一层结点个数，还是时间超限

<pre name="code" class="cpp">int countNodes(TreeNode* root)
{
	queue<TreeNode*> nodes;
	if(NULL==root)
		return 0;
	nodes.push(root);
	int depth(1);
	int result(0);
	while(!nodes.empty())
	{
		int length=nodes.size();
		int i=0;
		int k(0);
		while(i<length)
		{
			TreeNode* tmpNode=nodes.front();
			nodes.pop();
			if(tmpNode->left)
			{
				nodes.push(tmpNode->left);
				k++;
			}
			else
				break;
			if(tmpNode->right)
			{
				nodes.push(tmpNode->right);
				k++;
			}
			else
				break;
			
			i++;
		}
		if(i<length)
		{
			result=(1<<depth)-1+k;
			break;
		}
		depth++;
	}
	return result;
}

代码三：在代码一上的扩展，在网上看别人的。因为是完全二叉树，所以有且仅有最后一层和倒数第二层有叶子结点。首先统计最左边和最右边的深度，如果深度一致，说明是满二叉树（注意题目条件），则可以直接求出；如果不一致，则分别求左右孩子结点的结点个数。

<pre name="code" class="cpp">/**
 * 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(NULL==root)
    		return 0;
    	int height1(0),height2(0);
    	TreeNode* p=root;
    	while(p)
    	{
    		p=p->left;
    		height1++;
    	}
    	TreeNode* q=root;
    	while(q)
    	{
    		q=q->right;
    		height2++;
    	}
    	if(height1==height2)
    		return (1<<height1)-1;
    	return 1+countNodes(root->left)+countNodes(root->right);
    }
};