二叉樹中的最小路徑總和

Question

如何在二叉樹中找到最小路徑總和並打印路徑？ 路徑可以是從ROOT節點到任何LEAF節點的路徑。 我已經編寫了C ++代碼來找到最小和，但是在打印路徑時遇到了問題。

int MinTreePathSum(TreeNode *head, vector<TreeNode> &path)
{
    if(!head)  // head is NULL
        return 0;
    else if(!(head->left) && head->right)  // only head->left is NULL
        return head->val+MinTreePathSum(head->right, path);
    else if(!(head->right) && head->left)  // only head->right is NULL
        return head->val+MinTreePathSum(head->left, path);
    else
        return head->val+min(MinTreePathSum(head->left, path), MinTreePathSum(head->right, path));  // none of left and right are NULL
}

沒有使用參數列表中的path ，有人可以幫我打印路徑總和最小的路徑嗎？

Answer 1

而不是return head->val+min(MinTreePathSum(head->left, path), MinTreePathSum(head->right, path)); 檢查左右路徑中的哪一條是最小的。 通過保存它們，您可以找到路徑。

int MinTreePathSum(TreeNode *head, string &path)
{
    if(!head)  // head is NULL
        return 0;
    else if(!(head->left) && head->right)  // only head->left is NULL
    {   
        string p; 
        int val = head->val+MinTreePathSum(head->right, p);
        path = "R" + p;
        return val;
    }
    else if(!(head->right) && head->left)  // only head->right is NULL
    {
        string p;
        int val = head->val+MinTreePathSum(head->left, p);
        path = "L" + p;
        return val;
    }
    else
    {
        int vl,vr,val;
        string pl,pr;
        vl = MinTreePathSum(head->left, pl);
        vr = MinTreePathSum(head->right, pr);
        if ( vl < vr ){
             val = vl;
             path = "L" + pl;
        } else {
             val = vr;
             path = "R" + pr;
        }
        return head->val + val;
    }
}

Answer 2

我認為您真的很近...只要將當前節點添加到路徑中，如果當前節點有兩個子節點，則選擇最短路徑：

int MinTreePathSum(TreeNode *head, vector<TreeNode> &path)
{
    if(!head)  // head is NULL
        return 0;
    else if(!(head->left) && head->right) {  // only head->left is NULL
        path.push(*head);
        return head->val+MinTreePathSum(head->right, path);
    }else if(!(head->right) && head->left) { // only head->right is NULL
        path.push(*head);
        return head->val+MinTreePathSum(head->left, path);
    }else{ // two children, must choose one...
        path.push(*head);

        // get left and right paths
        pathLeft = vector<TreeNode>(); 
        pathRight = vector<TreeNode>();
        int valLeft = MinTreePathSum(head->left, pathLeft);
        int valRight = MinTreePathSum(head->right, pathRight);

        // actually copy the shortest path
        if (valLeft < valRight) {
            for(int i = 0; i < pathLeft.size(); ++i) {
                path.push(pathLeft[i]);
            }
        }else{
            for(int i = 0; i < pathRight.size(); ++i) {
                path.push(pathRight[i]);
            }
        }  
        // finally return the minimum path, which is the one we put in "path" already
        return head->val + min(valLeft, valRight);
    }
}