Binary Tree Longest Consecutive Sequence

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.

Tips:

简单的DFS，重点是local_max,即先算出这次遍历的最大长度，然后再和总最大长度进行比较。

Code：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int longestConsecutive(TreeNode root) {
        int result = dfs(root, -1, 0);
        return result;
    }
    private int dfs(TreeNode root, int prev, int local_max) {
        if (root == null) {
            return local_max;
        }
        if (root.val != prev + 1) {
            local_max = 1;
        } else {
            local_max++;
        }
        int left = dfs(root.left, root.val, local_max);
        int right = dfs(root.right, root.val, local_max);
        int result = Math.max(local_max, Math.max(left, right));
        return result;
    }
}