LeetCode OJ 之 Count Complete Tree Nodes（计算完全二叉树的结点数目）

来源：互联网发布：tortoisesvn for mac 编辑：程序博客网时间：2024/04/20 20:33

题目：

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.

思路：

最简单的思路：return 1 + countNodes(root->left) + countNodes(root->right);

可惜超时了。由于是完全二叉树，根结点的左右子树至少有一个是满二叉树，因此可以在递归之前先算出满二叉树的结点数目。

代码1：

/** * 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 height = 0;        TreeNode *leftNode = root , *rightNode = root;        while(leftNode && rightNode)        {            leftNode = leftNode->left;            rightNode = rightNode->right;            height++;        }        //如果leftNode为空，说明是满二叉树，返回2^h-1        if(leftNode == NULL)            return (1 << height) - 1;//这里一定要加（），因为-的优先级高于移位        else            return 1 + countNodes(root->left) + countNodes(root->right);    }    };

代码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 height(TreeNode *root)     {        if(root == NULL)            return -1;  //叶结点的子节点返回-1，这把叶结点那一行抵消掉了，故计算的高度不包括最后一行        else            return 1 + height(root->left);    }    int countNodes(TreeNode *root)     {        int h = height(root);        if(h < 0)            return 0;        if(height(root->right) == h-1)            return (1 << h) + countNodes(root->right);//如果root->right返回h-1，则说明最后一行的最后一个结点位于右子树，即说明左子树是满二叉树，左子树的结点数目即为2^h-1，再加上根结点就是2^h,然后+右子树的结点数        else            return (1 << h-1) + countNodes(root->left);//如果root->right返回的<h-1，则说明最后一行的最后一个结点位于左子树，即说明右子树的满二叉树，右子树的结点数目即为2^(h-1)-1，再加上根结点就是2^(h-1)，然后+左子树的结点数    }};

代码3：

/** * 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 height(TreeNode *root)     {        if(root == NULL)            return -1;  //叶结点的子节点返回-1，这把叶结点那一行抵消掉了，故计算的高度不包括最后一行        else            return 1 + height(root->left);    }    int countNodes(TreeNode *root)     {        int result = 0;        int h = height(root);        if(h < 0)            return 0;        while(root)        {            if(height(root->right) == h-1)            {                result += 1<<h;                root = root->right;            }            else            {                result += 1<<h-1;                root = root->left;            }            h--;        }        return result;            }};

0 0