leetcode 687. Longest Univalue Path 二叉树最长相等路径+ 深度优先遍历DFS

刺骨的言语ヽ痛彻心扉 2022-06-03 08:36 178阅读 0赞

Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root.

Note: The length of path between two nodes is represented by the number of edges between them.

Example 1:

Input:

5
             / \
            4   5
           / \   \
          1   1   5

Output:

2  
Example 2:

Input:

1
             / \
            4   5
           / \   \
          4   4   5

Output:

2  
Note: The given binary tree has not more than 10000 nodes. The height of the tree is not more than 1000.

本题题意很简单，属于写两个递归函数相互调用的类型，这个值得学习和思考

建议和[leetcode 437. Path Sum III 深度优先遍历DFS][leetcode 437. Path Sum III _DFS] 和 [leetcode 124. Binary Tree Maximum Path Sum 最大路径和 + DFS深度优先搜索][leetcode 124. Binary Tree Maximum Path Sum _ _ DFS] 一起学习

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    
    /*
    struct TreeNode
    {
         int val;
         TreeNode *left;
         TreeNode *right;
         TreeNode(int x) : val(x), left(NULL), right(NULL) {}
    };
    */
    
    class Solution 
    {
    public:
        int res = 0;
        int longestUnivaluePath(TreeNode* root) 
        {
            if (root == NULL)
                return 0;
            else
            {
                int all = dfs(root->left, root->val) + dfs(root->right, root->val);
                int left = longestUnivaluePath(root->left);
                int right = longestUnivaluePath(root->right);
                return max(all, max(left, right));
            }
        }
    
        int dfs(TreeNode* root, int parent)
        {
            if (root == NULL || root->val !=parent)
                return 0;
            else
            {
                int left = 1 + dfs(root->left, parent);
                int right = 1 + dfs(root->right, parent);
                return max(left, right);
            }
        }
    };

[leetcode 437. Path Sum III _DFS]: http://blog.csdn.net/jackzhang_123/article/details/78752985
[leetcode 124. Binary Tree Maximum Path Sum _ _ DFS]: http://blog.csdn.net/jackzhang_123/article/details/77989142