package com.fishercoder.solutions;

import com.fishercoder.common.classes.TreeNode;

/**
 * Given a binary tree, find the length of the longest consecutive sequence path.

 The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections.
 The longest consecutive path need to be from parent to child (cannot be the reverse).

 For example,
   1
   \
   3
  / \
 2  4
     \
     5
 Longest consecutive sequence path is 3-4-5, so return 3.

   2
    \
    3
   /
  2
 /
1
 Longest consecutive sequence path is 2-3,not3-2-1, so return 2.
 */
public class _298 {

    private int max = 1;

    public int longestConsecutive(TreeNode root) {
        if (root == null) {
            return 0;
        }
        dfs(root, 0, root.val);
        return max;
    }

    private void dfs(TreeNode root, int curr, int target) {
        if (root == null) {
            return;
        }
        if (root.val == target) {
            curr++;
        } else {
            curr = 1;
        }
        max = Math.max(max, curr);
        dfs(root.left, curr, root.val + 1);
        dfs(root.right, curr, root.val + 1);
    }

}

package com.fishercoder.solutions;

import com.fishercoder.common.classes.TreeNode;

/**
 * Given a binary tree, find the length of the longest consecutive sequence path.

 The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections.
 The longest consecutive path need to be from parent to child (cannot be the reverse).

 For example,
   1
   \
   3
  / \
 2  4
     \
     5
 Longest consecutive sequence path is 3-4-5, so return 3.

   2
    \
    3
   /
  2
 /
1
 Longest consecutive sequence path is 2-3,not3-2-1, so return 2.
 */
public class _298 {

    public static class Solution1 {
        private int max = 1;

        public int longestConsecutive(TreeNode root) {
            if (root == null) {
                return 0;
            }
            dfs(root, 0, root.val);
            return max;
        }

        private void dfs(TreeNode root, int curr, int target) {
            if (root == null) {
                return;
            }
            if (root.val == target) {
                curr++;
            } else {
                curr = 1;
            }
            max = Math.max(max, curr);
            dfs(root.left, curr, root.val + 1);
            dfs(root.right, curr, root.val + 1);
        }
    }
}

refactor 298