[LeetCode] Count Complete Tree Nodes

标签：leetcode   c++   

Count Complete Tree Nodes

 

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.

解题思路：

题意为计算完全二叉树的节点数目。完全二叉树是指除了最底层的节点之外的节点都有两个孩子节点，且最底层的节点尽量靠左的二叉树。

解法1：暴力遍历法，用递归一个个遍历。这种方法的时间复杂度为O(N)，会产生超时错误。

/**
 * 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==NULL){
            return 0;
        }
        return 1 + countNodes(root->left) + countNodes(root->right);
    }
};

注意一个小技巧，2^n次方可以通过移位来获得。

该方法的时间复杂度为O(h^2)，因为最耗时的操作是计算以某个节点为根节点的树的高度，最差情况下扫描次数为（1+2+...+h）*2。

/**
 * 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==NULL){
            return 0;
        }
        int leftH = getLeftHeight(root);
        int rightH = getRightHeight(root);
        if(leftH == rightH){
            return ( (1<<leftH) - 1);
        }
        return 1 + countNodes(root->left) + countNodes(root->right);
    }
    
    int getLeftHeight(TreeNode* root){
        int h = 0;
        while(root!=NULL){
            h++;
            root = root->left;
        }
        return h;
    }
    
    int getRightHeight(TreeNode* root){
        int h = 0;
        while(root!=NULL){
            h++;
            root = root->right;
        }
        return h;
    }
};

[LeetCode] Count Complete Tree Nodes

标签：leetcode   c++   

原文地址：http://blog.csdn.net/kangrydotnet/article/details/46521507