【Leetcode】124. Binary Tree Maximum Path Sum（二叉树最大路径）

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Given a non-empty 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.

Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

题目大意：

我们需要计算出二叉树当中取值最大的一个路径，题目当中如此描述树中的路径，该路径一定通过父节点同时到左右两个子树的节点只能有一个。

eg

解题思路：

找出当前节点左右两个侧的最大值，作为父节点的传入值。同时如果满足了有（左中右）存在最大值，则比较当前的最大值。

class Solution {
private:
    int ans;
    int helper(TreeNode *tmp){
        
        if(tmp->left==NULL&&tmp->right==NULL){
            return tmp->val;
        }
        
        int l, r;
        int state_l = false, state_r = false;
        if(tmp->left){
            l = helper(tmp->left);
            state_l = true;
        }else{
            l = 0;
        }
        
        if(tmp->right){
            r = helper(tmp->right);
            state_r = true;
        }else{
            r = 0;
        }
        
        int max_num = tmp->val;
        
        if(tmp->val+l+r > tmp->val+l && tmp->val+l+r > tmp->val+r){
            if(tmp->val+l+r > ans){
                ans = tmp->val+l+r;
            }
        }
        if(r > max_num && state_r){
            if(r > ans){
                ans = r;
            }
        }
        if(l > max_num && state_l){
            if(l > ans){
                ans = l;
            }
        }
        
        if(tmp->val+l > max_num && state_l){
            max_num = tmp->val + l;
        }
        if(tmp->val+r > max_num && state_r){
            max_num = tmp->val + r;
        }
        

        return max_num;
    }
public:
    int maxPathSum(TreeNode* root) { 
        ans = root->val;
        return max(helper(root), ans);
    }
};