687 Longest Univalue Path

687. Longest Univalue Path

1. Question

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.

2. Implementation

(1) DFS + Recursion

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int longestUnivaluePath(TreeNode root) {
        if (root == null) {
            return 0;
        }

        int[] res = new int[1];
        getLongestUnivaluePath(root, root.val, res);
        return res[0];
    }

    public int getLongestUnivaluePath(TreeNode node, int val, int[] res) {
        if (node == null) {
            return 0;
        }

        int leftPath = getLongestUnivaluePath(node.left, node.val, res);
        int rightPath = getLongestUnivaluePath(node.right, node.val, res);

        res[0] = Math.max(res[0], leftPath + rightPath);
        return val == node.val ? Math.max(leftPath, rightPath) + 1 : 0;
    }
}

3. Time & Space Complexity

DFS + Recursion: 时间复杂度O(n), 空间复杂度O(n)