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[LeetCode] Count Complete Tree Nodes

时间:2015-06-06 15:01:47      阅读:125      评论:0      收藏:0      [点我收藏+]

标签:leetcode

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.

解题思路

完全遍历会超时。可以从根节点开始分别计算左子树和右子树的高度,如果相等,则为满二叉树,结点个数为2n?1;如果高度不相等,则结点个数为左子树结点个数+右子树结点个数+1。(高度n是包含根节点的高度)

实现代码

// Runtime: 100 ms
/**
 * 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 leftHight = getLeft(root);
        int rightHeight = getRight(root);
        if (leftHight == rightHeight)
        {
            return pow(2, leftHight) - 1;
        }

        return 1 + countNodes(root->left) + countNodes(root->right);
    }

private:
    int getLeft(TreeNode *root)
    {
        int height = 0;
        while (root)
        {
            height++;
            root = root->left;
        }

        return height;
    }

    int getRight(TreeNode *root)
    {
        int height = 0;
        while (root)
        {
            height++;
            root = root->right;
        }

        return height;
    }
};

[LeetCode] Count Complete Tree Nodes

标签:leetcode

原文地址:http://blog.csdn.net/foreverling/article/details/46387781

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