124.leetcode Binary Tree Maximum Path Sum(hard)[先序遍历]

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.

要求从一个节点到另一个节点的最大和值，那么对于每个节点获取通过该节点得到的最大的值= root->val+左边最大和值+左边最大和值，如果和值都<0那么就只有当前节点会最大。getSum返回的是经过该节点的一条单路径的最大值，不能同时加上左右两边的值，因为这个值作为返回值上层才是转折点。

class Solution {
public:
    int max;
    int getmax(int a,int b)
    {
        if(a>=b)
          return a;
        else
          return b;
    }
    int getSum(TreeNode* root)
    {
        int value = root->val;
        int lmax = 0,rmax = 0;
        if(root->left != NULL)
        {
            lmax = getSum(root->left);
            if(lmax>0) value += lmax;
        }
        if(root->right != NULL)
        {
            rmax = getSum(root->right);
            if(rmax>0) value += rmax; 
        }
        if(value>max) max = value;
        return getmax(root->val,getmax(root->val+lmax,root->val+rmax));
        
    }
    int maxPathSum(TreeNode* root) {
        if(root == NULL) return 0;
        max = root->val;
        getSum(root);
        return max;
    }
};