[leetcode] 543. Diameter of Binary Tree
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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 thelongest path 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.
这道题是计算二叉树“直径”,题目难度为Easy。
这道题和第124题非常相近,感兴趣的同学可以顺便看下第124题(传送门)。
所谓的“直径”,即任意两节点之间的最大距离。“直径”所在的路径必定存在一个自己的根节点将其分为左右两部分,两部分分别是其子节点最大深度对应的节点路径,这样题目就转换为计算节点深度了。知道了两个子节点的深度之后就可以计算出以此节点为根的“直径”,通过深度优先遍历二叉树即可得出最终最大的“直径”。具体代码:
class Solution { int getRadius(TreeNode* n, int& diameter) { int lRadius = 0, rRadius = 0; if(n->left) lRadius = getRadius(n->left, diameter); if(n->right) rRadius = getRadius(n->right, diameter); diameter = max(diameter, lRadius+rRadius); return max(lRadius, rRadius) + 1; }public: int diameterOfBinaryTree(TreeNode* root) { if(!root) return 0; int diameter = INT_MIN; getRadius(root, diameter); return diameter; }};
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