Count Complete Tree Nodes -- leetcode
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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.
基本思路:折半
判断当前子树是否为满二叉树,如果是,则应用公式,根据高度求结点。2 ^ (h+1) - 1, h 从0 开始,即根为0层
如果不是满二叉树,则递归求出左子树和右子树的结点树。
判断是否为满二叉树,是通过从左子树往下遍历求高度,和通过右子树往下求高度,最后作比较完成的。
每次折半时,左子树和右子树中,必有一个是满二叉树。故可每次完成一半结点的统计。
时间复杂度为O(log(n)^2)
在leetcode上实际执行时间为128ms。
/** * 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) return 0; int level = -1; TreeNode *left = root; TreeNode *right = root; while (right) { ++level; left = left->left; right = right->right; } if (!left) return (1 << level+1) - 1; else return 1 + countNodes(root->left) + countNodes(root->right); }};
算法二,折半
算法一中,每次递归时,都会进行两次求高度运算。
此算法将省去一次高度运算。在进行子树细分时,高度是逐渐递减的,因此左边子树的高度无须每次进行运算。
但是思路与算法一略有差别。
1. 求出右子树的高度。
2. 如果此高度与左子树的高度一致。 则说明左子树是,一棵满二叉树。左子树的节点数可以用公式求解。只需要再对右子树继续进行递归折半。
3. 如果此高度与左子树的高度不一致。 则说明右子树,是一棵满二叉树。右子树的节点数可以用公式求解。然后再对左子树进行递归折半。
在leetcode上实际执行时间为96ms。
class Solution {public: int countNodes(TreeNode* root) { return count(root, height(root)); } int count(TreeNode* root, int level) { if (!root) return 0; int rightH = height(root->right); if (rightH == level-1) return (1 << level) + count(root->right, level-1); else return (1 << level-1) + count(root->left, level-1); } int height(TreeNode* root) { int level = -1; while (root) { ++level; root = root->left; } return level; }};
最后附上自己的第一版算法,留作纪念
在leetcode上实际执行时间为120ms。
基本思路,基于折半。
同时,关注点,在于对最后一层的节点数的统计。代码中,count是统计最后一层的节点数。
每一层递归中,统计当前子树,最下节点,是否达到指定的高度。
class Solution {public: int countNodes(TreeNode* root) { if (!root) return 0; if (!root->left) return 1; int total = 0; int nodes = 1; int level = 1; TreeNode *p = root; while (p->left) { total += nodes; p = p->left; nodes <<= 1; ++level; } return total + count(root, nodes, level); } int count(TreeNode* root, int last_level_nodes, int level) { if (is_complete(root, level)) return last_level_nodes; last_level_nodes >>= 1; --level; int total = 0; if (root->left) total = count(root->left, last_level_nodes, level); if (total == last_level_nodes && root->right) total += count(root->right, last_level_nodes, level); return total; } bool is_complete(TreeNode* root, int level) { while (root && level) { root = root->right; --level; } return !root && !level; }};
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