124. Binary Tree Maximum Path Sum (Tree; DFS)

標籤：

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 does not need to go through the root.

For example:
Given the below binary tree,

       1      /      2   3

Return 6.

思路：存在val小於零的情況，所以path不一定是從葉子節點到葉子節點；

/** * 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 maxPathSum(TreeNode *root) {        maxSum = INT_MIN;        calRootSum(root);                return maxSum;            }        int calRootSum(TreeNode *root)    {        int leftSum = 0;        int rightSum = 0;        int sum;                if(root->left)         {            leftSum = calRootSum(root->left); //traverse left subtree        }        if(root->right)         {            rightSum = calRootSum(root->right);//traverse right subtree        }                //val小於零的情況        if(leftSum < 0)         {            leftSum = 0;        }        if(rightSum < 0)        {            rightSum = 0;        }                sum = root->val+leftSum+rightSum;        if(sum > maxSum)        {            maxSum = sum;        }                sum = root->val+max(leftSum,rightSum);        return sum;    }private:    int maxSum;};

 

124. Binary Tree Maximum Path Sum (Tree; DFS)