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
postorder traversal with recursion
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int res = 0;
pair<int, int> postorder(TreeNode * root) {
if (!root) return {0, 0};
auto [linc, ldec] = postorder(root->left);
auto [rinc, rdec] = postorder(root->right);
int lval = root->left ? root->left->val : INT_MIN;
int rval = root->right ? root->right->val : INT_MIN;
int inc, dec, val = root->val;
inc = max(val + 1 == lval ? linc : 0, val + 1 == rval ? rinc : 0) + 1;
dec = max(val - 1 == lval ? ldec : 0, val - 1 == rval ? rdec : 0) + 1;
// if two subtree are both inc, then inc will be the largest and dec will be 1
// if one subtree is in can another is dec, thus formula also works
res = max(res, inc + dec - 1);
return {inc, dec};
}
int longestConsecutive(TreeNode* root) {
if (root) postorder(root);
return res;
}
};