Lintcode 614 · Binary Tree Longest Consecutive Sequence II (二叉树后序遍历好题)

614 · Binary Tree Longest Consecutive Sequence II
Algorithms
Medium

Description
Given a binary tree, find the length(number of nodes) of the longest consecutive sequence(Monotonic and adjacent node values differ by 1) path.
The path could be start and end at any node in the tree

Example
Example 1:

Input:
{1,2,0,3}
Output:
4
Explanation:
1
/
2 0
/
3
0-1-2-3
Example 2:

Input:
{3,2,2}
Output:
2
Explanation:
3
/
2 2
2-3
Tags
Company
Google
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解法1：后序遍历。helper()返回当前节点的incrLen 和 decrLen，同时维护当前最大的maxLen = max(maxLen, incrLen + decrLen - 1)。

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: the root of binary tree
     * @return: the length of the longest consecutive sequence path
     */
    int longestConsecutive2(TreeNode *root) {
        if (!root) return 0;
        helper(root);
        return maxLen;
    }
private:
    int maxLen = 1;
    vector<int> helper(TreeNode *root) {
        if (!root) return {};
        int incrLen = 1, decrLen = 1;
        vector<int> left = helper(root->left); //<incrLen, decrLen>
        vector<int> right = helper(root->right); //<incrLen, decrLen>
        if (root->left && root->val == root->left->val + 1) {
            incrLen = left[0] + 1;
        } 
        if (root->left && root->val == root->left->val - 1) {
            decrLen = left[1] + 1;
        }

        if (root->right && root->val == root->right->val + 1) {
            incrLen = max(incrLen, right[0] + 1);
        }
        if (root->right && root->val == root->right->val - 1) {
            decrLen = max(decrLen, right[1] + 1);
        }
        maxLen = max(maxLen, incrLen + decrLen - 1);
        return {incrLen, decrLen};
    }
};