Question

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.

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

Difficulty:Easy

Category:Tree, Recursion

Analyze

This problem: You need to find the longest path which each node have the same value in this tree.(The path can pass the root node.) The output value not the node number. It is the side numbers.

You can use the left len and right len and the root node together.

Output 3 is 2, not 3.

Solution

Solution 1: dfs

Time complexity: O(n) Space complexity: O(n)

Runtime: 140 ms, faster than 93.41% of C++ online submissions for Longest Univalue Path. Memory Usage: 50.2 MB, less than 47.32% of C++ online submissions for Longest Univalue Path.

class Solution {
 public:
  int longestUnivaluePath(TreeNode* root) {
    int max_len = 0;
    dfs(root, max_len);
    return max_len;
  }

 private:
  int dfs(TreeNode* root, int& max_len) {
    if (!root) return 0;
    int l = dfs(root->left, max_len);
    int r = dfs(root->right, max_len);
    int l_len = 0, r_len = 0;

    // The l is the long for the root->left, So, it need to return l_len
    if (root->left && root->left->val == root->val) l_len = l + 1;
    if (root->right && root->right->val == root->val) r_len = r + 1;

    // The path may pass the root node
    max_len = max(max_len, l_len + r_len);
    return max(l_len, r_len);
  }
};

其他类似解法可参考博客：Longest Univalue Path 最长相同值路径