刷题124. Binary Tree Maximum Path Sum

一、题目说明

题目124. Binary Tree Maximum Path Sum，给一个非空二叉树，求最大路径之和。

二、我的解答

这个题目，经过几个小时的思考，终于做出来了。一个树的最大路径，可能出现在“左子树”，“右子树”，或者包含“根节点”。其中dfs用来递归计算从根到叶的最大路径之和。

class Solution{
	public:
		int dfs(TreeNode* root){
			if(root==NULL) return INT_MIN;
			if(root->left==NULL && root->right==NULL){
				if(root->val>curMax) curMax = root->val; 
				return root->val;
			}

			int left = dfs(root->left);
			int right = dfs(root->right);
			
			//当左右子树最大路径之和为正，则可能产生最大路径
			int cur = root->val;
			if(left>0 && right>0){
				cur += left;
				cur += right;
				if(cur>curMax) curMax = cur;
			}
			
			cur = root->val;
			int m = max(left,right);
			if(m>0){
				cur += m;
			}
		
			if(cur>curMax) curMax = cur;
			return cur;
		}
		int maxPathSum(TreeNode* root){
			if(root==NULL) return INT_MIN;
			if(root->left==NULL && root->right==NULL){
				return root->val;
			}
			curMax = INT_MIN;
			if(root->val>curMax) curMax = root->val;

			int left = dfs(root->left);
			int right = dfs(root->right);
			
			int cur = root->val;
			if(left>0 && right>0){
				cur += left;
				cur += right;
			}else if(left>0){
				cur += left;
			}else if(right>0){
				cur += right;
			}
			
			if(cur>curMax) curMax = cur;
			return curMax;
		}
	private:
	    int curMax;
};

性能如下：

Runtime: 28 ms, faster than 86.43% of C++ online submissions for Binary Tree Maximum Path Sum.
Memory Usage: 24.2 MB, less than 96.97% of C++ online submissions for Binary Tree Maximum Path Sum.

三、优化措施

无