549. Binary Tree Longest Consecutive Sequence II

最新推荐文章于 2020-12-05 14:44:06 发布

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最新推荐文章于 2020-12-05 14:44:06 发布

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原文链接：http://www.cnblogs.com/jxr041100/p/7885766.html

本文介绍了一种在二叉树中寻找最长连续路径的方法。连续路径可以递增或递减，且不必遵循父节点到子节点的顺序。通过深度优先搜索(DFS)策略实现这一目标，并提供了具体实例。

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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].

/**
 * 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 longestConsecutive(TreeNode* root) {
        int longestLen = 0;
        dfs(root,NULL,longestLen);
        return longestLen;
    }
private:
    pair<int,int> dfs(TreeNode *root,TreeNode*pre,int &longestLen)
     {
        if(!root) return {0,0};
        pair<int,int> res1,res2;
        pair<int,int> left = dfs(root->left,root,longestLen);
        pair<int,int> right = dfs(root->right,root,longestLen);
        longestLen = max(longestLen,left.first+right.second+1);
        longestLen = max(longestLen,left.second+right.first+1);
        int inc = 0, dec = 0;
        if ( pre && root->val == pre->val + 1 ) {
         inc = max(left.first, right.first) + 1;
        }
        if (pre&& root->val == pre->val - 1 ) {
        dec = max(left.second, right.second) + 1;
        }
       return {inc,dec};
     }
};