549 Binary Tree Longest Consecutive Sequence II

Problem:

Given a binary tree, you need to find the length of Longest Consecutive Path in Binary Tree.

Especially, this path can be either increasing or decreasing. For example, [1,2,3,4] and [4,3,2,1] are both considered valid, but the path [1,2,4,3] is not valid. On the other hand, the path can be in the child-Parent-child order, where not necessarily be parent-child order.

Example 1:

``````Input:
1
/ \
2   3
Output: 2
Explanation: The longest consecutive path is [1, 2] or [2, 1].
``````

Example 2:

``````Input:
2
/ \
1   3
Output: 3
Explanation: The longest consecutive path is [1, 2, 3] or [3, 2, 1].
``````

Note: All the values of tree nodes are in the range of [-1e7, 1e7].

Solutions:

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private class Node {
private int incr;
private int decr;
public Node() {
incr = 0;
decr = 0;
}
}
public int longestConsecutive(TreeNode root) {
int[] res = new int[1];
res[0] = 0;
process(root, res);
return res[0];
}
private Node process(TreeNode node, int[] res) {
if (node == null) {
return new Node();
}
Node left = process(node.left, res);
Node right = process(node.right, res);
Node curr = new Node();
int sum_incr = 1;
int sum_decr = 1;
if (node.left != null) {
if (node.left.val == node.val - 1) {
curr.decr = Math.max(curr.decr, left.decr + 1);
sum_incr += left.decr + 1;
}
if (node.left.val == node.val + 1) {
curr.incr = Math.max(curr.incr, left.incr + 1);
sum_decr += left.incr + 1;
}
}
if (node.right != null) {
if (node.right.val == node.val - 1) {
curr.decr = Math.max(curr.decr, right.decr + 1);
sum_decr += right.decr + 1;
}
if (node.right.val == node.val + 1) {
curr.incr = Math.max(curr.incr, right.incr + 1);
sum_incr += right.incr + 1;
}
}
res[0] = Math.max(res[0], Math.max(sum_incr, sum_decr));
return curr;
}
}
``````