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].
Solution:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
int maxval = 0;
public int longestConsecutive(TreeNode root) {
longestPath(root);
return maxval;
}
public int[] longestPath(TreeNode root) {
if (root == null) {
return new int[] {0, 0};
}
int inr = 1, dcr = 1;
if (root.left != null) {
int[] l = longestPath(root.left);
if (root.val == root.left.val + 1) {
dcr = l[1] + 1;
}
else if (root.val == root.left.val - 1) {
inr = l[0] + 1;
}
}
if (root.right != null) {
int[] r = longestPath(root.right);
if (root.val == root.right.val + 1) {
dcr = Math.max(dcr, r[1] + 1);
}
else if (root.val == root.right.val - 1) {
inr = Math.max(inr, r[0] + 1);
}
}
maxval = Math.max(maxval, dcr + inr - 1);
return new int[] {inr, dcr};
}
}