[LintCode] 619 Binary Tree Longest Consecutive Sequence III 解题报告

Description
It's follow up problem for Binary Tree Longest Consecutive Sequence II

Given a k-ary tree, find the length of the longest consecutive sequence path.
The path could be start and end at any node in the tree



Example
An example of test data: k-ary tree 5<6<7<>,5<>,8<>>,4<3<>,5<>,3<>>>, denote the following structure:


     5
   /  
  6     4
 /|   /|
7 5 8 3 5 3

Return 5, // 3-4-5-6-7

5/10/2017

算法班，这道题其实已经不是binary tree了，不过至少是从那里演化而来。

未经验证，注意第44行，3方面比较

起始值为0还是1跟第44行+1还是-1有关。

/**
 * Definition for a multi tree node.
 * public class MultiTreeNode {
 *     int val;
 *     List<TreeNode> children;
 *     MultiTreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
     * @param root the root of k-ary tree
     * @return the length of the longest consecutive sequence path
     */
    class Result {
        int maxLength;
        int maxUp;
        int maxDown;
        Result(int maxLength, int maxUp, int maxDown) {
            this.maxLength = maxLength;
            this.maxUp = maxUp;
            this.maxDown = maxDown;
        }
    }
    public int longestConsecutive3(MultiTreeNode root) {
        // Write your code here
        if (root == null) return 0;
        Result ret = longestPath(root);
        return ret.maxLength;
    }

    private void longestPath(MultiTreeNode root) {
        if (root == null) {
            return new Result(0, 0, 0);
        }
        int maxLength = 0, maxUp = 0, maxDown = 0;
        for (TreeNode child: root.children) {
            Result tmp = longestPath(child);
            if (child.val - 1 == root.val) {
                maxUp = Math.max(maxUp, tmp.maxUp + 1);
            }
            if (child.va + 1 == root.val) {
                maxDown = Math.max(maxDown, tmp.maxDown + 1);
            }
            maxLength = Math.max(Math.max(maxLength, tmp.maxLength), maxUp + maxDown + 1);
        }
        return new Result(maxLength, maxUp, maxDown);
    }

}