543. Diameter of Binary Tree (二叉树的直径)
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Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longestpath between any two nodes in a tree. This path may or may not pass through the root.
Example:
Given a binary tree
1 / \ 2 3 / \ 4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
Note: The length of path between two nodes is represented by the number of edges between them.
这道题让我们求二叉树的直径,并告诉了我们直径就是两点之间的最远距离,根据题目中的例子也不难理解题意。我们再来仔细观察例子中的那两个最长路径[4,2,1,3] 和 [5,2,1,3],我们转换一种角度来看,是不是其实就是根结点1的左右两个子树的深度之和再加1呢。那么我们只要对每一个结点求出其左右子树深度之和,再加上1就可以更新结果res了。为了减少重复计算,我们用哈希表建立每个结点和其深度之间的映射,这样某个结点的深度之前计算过了,就不用再次计算了,参见代码如下:
解法一:
class Solution {public: int diameterOfBinaryTree(TreeNode* root) { if (!root) return 0; int res = getHeight(root->left) + getHeight(root->right); return max(res, max(diameterOfBinaryTree(root->left), diameterOfBinaryTree(root->right))); } int getHeight(TreeNode* node) { if (!node) return 0; if (m.count(node)) return m[node]; int h = 1 + max(getHeight(node->left), getHeight(node->right)); return m[node] = h; }private: unordered_map<TreeNode*, int> m;};
上面的方法貌似有两个递归函数,其实我们只需要用一个递归函数就可以了,我们再求深度的递归函数中顺便就把直径算出来了,而且貌似不用进行优化也能通过OJ,参见代码如下:
解法二:
class Solution {public: int diameterOfBinaryTree(TreeNode* root) { int res = 0; maxDepth(root, res); return res; } int maxDepth(TreeNode* node, int& res) { if (!node) return 0; int left = maxDepth(node->left, res); int right = maxDepth(node->right, res); res = max(res, left + right); return max(left, right) + 1; }};
虽说不用进行优化也能通过OJ,但是毕竟还是优化一下好一点啊,参见代码如下:
解法三:
class Solution {public: int diameterOfBinaryTree(TreeNode* root) { int res = 0; maxDepth(root, res); return res; } int maxDepth(TreeNode* node, int& res) { if (!node) return 0; if (m.count(node)) return m[node]; int left = maxDepth(node->left, res); int right = maxDepth(node->right, res); res = max(res, left + right); return m[node] = (max(left, right) + 1); }private: unordered_map<TreeNode*, int> m;};
笔记:递归调用,使用哈希表优化结构。
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