Binary Tree Maximum Path Sum

最新推荐文章于 2024-03-25 21:39:45 发布

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本文介绍了一种寻找给定二叉树中最大路径和的算法实现。该路径可以始于并结束于任意节点，通过递归遍历左子树和右子树，计算每个节点的最大贡献值和包含该节点的最大路径和。

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Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
   
    int maxPathSum(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int max_sum = INT_MIN;
        return max(maxSum(root,max_sum),max_sum);
    }
    
    int maxSum(TreeNode* root,int& max_sum)
    {
       
        if(root == NULL) return 0;
        int left_sum = maxSum(root->left,max_sum);
        int right_sum = maxSum(root->right,max_sum);
        
        int sum1 = root->val + max(0, max(left_sum,right_sum));
        int sum2 = root->val + left_sum + right_sum;
        
        max_sum = max(max_sum, max(sum1,sum2));

        return sum1;
    }
};

144 milli secs