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.

这题是连续递增序列就行，
solution_extra是考虑续递递增|续递递减的情况，
另外因为本题是parent-child，所以pre-order post-order都可以。

Solution：DFS

思路：dfs 并传递 当前最大c_max_len_inc 和 上一个结点来做判断从而更新当前的c_max_len_inc，更新global result。

Time Complexity: O(N) Space Complexity: O(N) 递归缓存

Solution Code：
//parent2child连续递增序列

class Solution {
    private int result;
    public int longestConsecutive(TreeNode root) {
        if(root == null) return 0;
        result = 1;
        dfsHelper(root, null, 1);
        return result;
    }
    
    private void dfsHelper(TreeNode node, TreeNode prev, int c_max_len_inc) {
        if(node == null) return;
        if(prev != null && node.val == prev.val + 1) {
            c_max_len_inc++;
            if(c_max_len_inc > result) result = c_max_len_inc;
        }
        else {
            c_max_len_inc = 1;
        }
        dfsHelper(node.left, node, c_max_len_inc);
        dfsHelper(node.right, node, c_max_len_inc);
    }
}

Extra: parent2child可以连续递增|连续递减序列

//传递也应该写成package比较clear
class Solution_extra {
    private int result;
    public int longestConsecutive(TreeNode root) {
        if(root == null) return 0;
        result = 1;
        dfsHelper(root, null, 1, 1);
        return result;
    }
    
    private void dfsHelper(TreeNode node, TreeNode prev, int c_max_len_inc, int c_max_len_dec) {
        if(node == null) return;
        if(prev != null && node.val == prev.val + 1) {
            c_max_len_inc++;
            c_max_len_dec = 1;
            if(c_max_len_inc > result) result = c_max_len_inc;
        }
        else if(prev != null && node.val == prev.val - 1) {
            c_max_len_dec++;
            c_max_len_inc = 1;
            if(c_max_len_dec > result) result = c_max_len_dec;
        }
        else {
            c_max_len_inc = 1;
            c_max_len_dec = 1;
        }
        dfsHelper(node.left, node, c_max_len_inc, c_max_len_dec);
        dfsHelper(node.right, node, c_max_len_inc, c_max_len_dec);
    }
}