PROBLEM:
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.
SOLVE:
class Solution {
public:
int longestUnivaluePath(TreeNode* root) {
int lup = 0;
if (root) dfs(root, lup);
return lup;
}
private:
int dfs(TreeNode* node, int& lup) {
//lup:以node爲根節點的樹中最長路徑
int l = node->left ? dfs(node->left, lup) : 0; //l:左子樹從根節點開始的最長路徑 //注意:不一定是左子樹的最長路徑 //lup:更新爲左子樹的最長路徑
int r = node->right ? dfs(node->right, lup) : 0; //r:右子樹從根節點開始的最長路徑 //注意:不一定是右子樹的最長路徑 //lup:更新爲左子樹及右子樹中的最長路徑
int resl = node->left && node->left->val == node->val ? l + 1 : 0;
int resr = node->right && node->right->val == node->val ? r + 1 : 0;
lup = max(lup, resl + resr); //lup:更新後爲以node爲根節點的樹中最長路徑 //注意:路徑是邊數,而不是節點數
return max(resl, resr); //從根節點開始的最長路徑
}
};