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;    }};