[Leetcode] 687. Longest Univalue Path 解题报告

题目：

Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root.

Note: The length of path between two nodes is represented by the number of edges between them.

Example 1:

Input:

              5
             / \
            4   5
           / \   \
          1   1   5

Output:

2

Example 2:

Input:

              1
             / \
            4   5
           / \   \
          4   4   5

Output:

2

Note: The given binary tree has not more than 10000 nodes. The height of the tree is not more than 1000.

思路：

符合条件的path可以分为两种类型：1）经过root的；2）不经过root的。经过root的，其必然是从左边的某个leaf一致延伸到root，然后再从root延伸到右边的某个leaf，所以我们定义一个新的私有函数longestSubPath来实现计算。不经过root的可以通过递归调用root的左右子树来实现。

代码：

/**
 * 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 longestUnivaluePath(TreeNode* root) {
        if (root == NULL) {
            return 0;
        }
        // type 1: the path that goes through the root
        int ans1 = 0;
        if (root->left && root->left->val == root->val) {
            ans1 += longestSubPath(root->left) + 1;
        }
        if (root->right && root->right->val == root->val) {
            ans1 += longestSubPath(root->right) + 1;
        }
        // type 2: the path that does not go through the root
        int ans2 = longestUnivaluePath(root->left);
        int ans3 = longestUnivaluePath(root->right);
        return max(ans1, max(ans2, ans3));
    }
private:
    int longestSubPath(TreeNode* root) {
        // get the longest sub path that goes from root to a leaf
        if (root == NULL) {
            return 0;
        }
        int ret = 0;
        if (root->left && root->left->val == root->val) {
            ret = longestSubPath(root->left) + 1;
        }
        if (root->right && root->right->val == root->val) {
            ret = max(ret, longestSubPath(root->right) + 1);
        }
        return ret;
    }
};