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.
Runtime: 72 ms, faster than 28.47% of C++ online submissions for Longest Univalue Path.
对于这种不经过root的求和题往往都需要一个临时变量，然后考虑一下根节点和子节点的关系，用一个引用得到最优解。
/**
* 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) {
int ret = , tmpret = ;
helper(root,tmpret, ret);
return ret == ? : ret - ;
}
int helper(TreeNode* root, int& tmpret, int& ret){
if(!root) return ;
int rl = helper(root->left, tmpret, ret);
int rr = helper(root->right, tmpret, ret);
if(!root->left && !root->right) {
tmpret = ;
//ret = 1;
return ;
} else if(!root->left && root->right){
if(root->val == root->right->val){
tmpret = max(tmpret, rr + );
ret = max(ret, tmpret);
return +rr;
} else return ;
} else if(root->left && !root->right){
if(root->val == root->left->val){
tmpret = max(tmpret, +rl);
ret = max(ret, tmpret);
return +rl;
} else return ;
} else {
if(root->val == root->left->val && root->val == root->right->val){
tmpret = max(tmpret, +rr + rl);
ret = max(ret, tmpret);
return +max(rr,rl);
} else if (root->val == root->left->val){
tmpret = max(tmpret, +rl);
ret = max(ret, tmpret);
return +rl;
} else if (root->val == root->right->val){
tmpret = max(tmpret, +rr);
ret = max(ret, tmpret);
return +rr;
} else return ;
}
}
};