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:
对于每一棵子树,我们可以可以选择左边最大,右边最大,或者左边+右边
而这些选择是在某种条件下才可能发生。比如
Input:
4
/
4 4
当左边和右边相等并且等于root时,我们才可能合并左右。 而在左边不能与root或则右边不等于root,我们只需要取最大的一个作为这棵子树的最大值了,代码如下
/**
* 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 ans;
int dfsmax(TreeNode* root) {
if (root == nullptr) return 0;
int l = dfsmax(root->left);
int r = dfsmax(root->right);
int m1 = 0;
int m2 = 0;
if (root->left == nullptr || root->left->val != root->val) l = 0;
else if (root->left->val == root->val) ++l, m1 = 1;
if (root->right == nullptr || root->right->val != root->val) r = 0;
else if (root->right->val == root->val) ++r, m2 = 1;
if (m1 && m2) ans = max(ans, l + r);
ans = max(ans, max(l, r));
return max(l, r);
}
int longestUnivaluePath(TreeNode* root) {
ans = 0;
dfsmax(root);
return ans;
}
};