Binary Tree Maximum Path Sum

Binary Tree Maximum Path Sum.png

解題思路 ：

此題反而比 II 還要多一個考慮的點 如果 left + right + root->val 本身的值就已經是最大值 必須要把此值保留 特別注意 findMax function 回傳的並不是 res 而是 node 本身的值 加上左邊或右邊二選一 ( 選最大者 因為一次只能走一條路訪問一個子節點 )

C++ code :

 /**
 
 
 
  
Definition of TreeNode:
 
  
class TreeNode {
 
  
public:
 
  
int val;
 
  
TreeNode *left, *right;
 
  
TreeNode(int val) {
 
  
    this->val = val;
 
  
    this->left = this->right = NULL;
 
  
}
 
  
}
 */
 
 
 
 
class Solution {
 
 
public:
 /**
 * @param root: The root of binary tree.
 * @return: An integer
 */
 // write your code here
 int findMax(TreeNode *root, int &res)
 {
 if(root == nullptr) return 0;
 int sum = root->val;
 int left = findMax(root->left, res);
 int right = findMax(root->right, res);
 if(left > 0) sum += left;
 if(right > 0) sum += right;
 res = max(res, sum);
 return max(max(left, right) + root->val, root->val);
 }
 
 
int maxPathSum(TreeNode *root) {
    int res = INT_MIN;
    findMax(root, res);
    return res;
}
 
 
};