函数参数m表示不包含本节点或者同时包含左右子节点的最大值,mc表示包含本节点的最大值,这样遍历至某个节点的最大值等于,左子树的m/mc值,或者右子树的m/mc值,或者max(0,左子树mc值) + max(0,右子树mc值) + 本节点值,取这几个中最大的
代码如下:

#define INF 1e9

class Solution {
public:

void dfs(TreeNode* root, int& m, int& mc){
  if(!root) return ;
  mc=root->val;
  m=-INF;
  int l,lc;
  l=lc=-INF;
  int r,rc;
  r=rc=-INF;
  if(root->left){
    dfs(root->left,l,lc);
  }
  if(root->right){
    dfs(root->right,r,rc);
  }
  if(l>-INF){
    if(r>-INF){
      m = max(max(max(l,r),lc+rc+mc), max(lc,rc));
      mc = mc + max(0,max(lc,rc));
    }else{
      if(rc>-INF){
        m = max(max(l,lc+rc+mc),max(rc,lc));
        mc = mc + max(0,max(lc,rc));
      }else{
        m = max(l, lc);
        mc = mc + max(0, lc);
      }
    }
  }else{
    if(lc>-INF){
      if(r>-INF){
        m = max(max(r, lc+rc+mc), max(lc,rc));
        mc = mc + max(0, max(lc,rc));
      }else{
        if(rc>-INF){
          m = max(lc + rc + mc, max(lc,rc));
          mc = mc + max(0, max(lc,rc));
        }else{
          m = lc;
          mc = mc + max(0,lc);
        }
      }
    }else{
      if(r>-INF){
        m = max(r,rc);
        mc = mc + max(0, rc);
      }else{
        if(rc>-INF){
          m = rc;
          mc = mc + max(0, rc);
        }
      }
    }
  }
}

    int maxPathSum(TreeNode* root) {
        // write code here
        int m, mc;
        m=0;
        mc=root->val;
        dfs(root,m,mc);
        return max(m,mc); 
    }
};