编程练习（第五周）

124. Binary Tree Maximum Path Sum

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Description Submissions Solutions

• Total Accepted: 90097
• Total Submissions: 354576
• Difficulty: Hard
• Contributors: Admin

Given a binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

自我分析：对于这种二叉树的问题很适合用递归的方法来实现，求最大的路径和，所以是考虑到负值的情况，路径是唯一的，所以只要对左右分别进行递归，找出唯一的最大路径和最后左右相加即可。

代码如下：

    int Path(TreeNode* root,int&maxSum)
    {
        if(root==NULL)
            return 0;
        int leftmax = Path(root->left,maxSum);
        int rightmax = Path(root->right,maxSum);
        maxSum = max(maxSum,max(0,leftmax)+max(0,rightmax)+root->val);
        return max(0,max(leftmax,rightmax))+root->val;
    }
    int maxPathSum(TreeNode *root) {
        int maxSum = INT_MIN;
        Path(root,maxSum);
        return maxSum;
    }