6

我正在努力解决这个问题，我想以非递归方式解决这个问题。我的算法似乎没有逻辑错误，73% 的测试用例通过了。但无法处理大数据，报“Time Limit Exceeded”。如果有人能给我一些提示，我很感激如何以非递归方式进行操作并避免超出时间限制，在此先感谢！

问题链接

我相信在 LeetCode 中也有类似的。

http://www.lintcode.com/en/problem/binary-tree-maximum-path-sum-ii/

问题描述：

给定一棵二叉树，找到从根开始的最大路径和。路径可以在树中的任何节点处结束，并且其中至少包含一个节点。

例子：

给定以下二叉树：

1

/ \

2 3

返回 4. (1->3)

法官

超过时间限制

总运行时间：1030 毫秒

输入 输入数据

{-790,-726,970,696,-266,-545,830,-866,669,-488,-122,260,116,521,-866,-480,-573,-926,88,733,#,#,483,-935,-285,-258,892,180,279 ,-935,675,2,596,5,50,830,-607,-212,663,25,-840,#,#,-333,754,#,817,842,-220,-269,9,-862,-78,-473,643,536,- 142,773,485,262,360,702,-661,244,-96,#,519,566,-893,-599,126,-314,160,358,159,#,#,-237,-522,-327,310,-506,462,-705,0,-945,-73,19,3 193,-205,-92,795,-99,-983,-658,-114,-706,987,292,#,234,-406,-993,-863,859,875,383,-729,-748,-258,329,431,-188,-375 ,-696,-856,825,-154,-398,-917,-70,105,819,-264,993,207,21,-102,50,569,-824,-604,895,-564,-361,110,-965,-11,557,#,202,213 ,-141,759,214,207,135,329,15,#,#,244,#,334,628,509,627,-737,-33,-339,-985,349,267,-505,-527,882,-352,-357,-630,782,23,-215,-555, 835,-421,751,0,-792,-575,-615,-690,718,248,882,-606,-53,157,750,862,#,940,160,47,-347,-101,-947,739,894,#,-658,-90,-277 ,-925,997,862,-481,-83,708,706,686,-542,485,517,-922,978,-464,-923,710,-691,168,-607,-888,-439,499,794,-601,435,-114,-337,422,#,-3,-859, 224,902,#,577,#,-386,272,-9 ...

预期的

6678

我的代码 C++

/**
 * 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
     */
    int maxPathSum2(TreeNode *root) {
        if (root == NULL) return 0;
        findLeaf(root);
        return global_max;
    }

private:
    int global_max = INT_MIN;

    void findLeaf(TreeNode* root) {
        unordered_map<TreeNode*, TreeNode*> parent;
        stack<TreeNode*> traverse;
        parent[root] = NULL;
        traverse.push(root);

        while(!traverse.empty()) {
            TreeNode* p = traverse.top();
            traverse.pop();
            if (!p->left && !p->right) {
                findPathMaxSum(p, parent);
            }
            if (p->right) {
                parent[p->right] = p;
                traverse.push(p->right);
            }
            if (p->left) {
                parent[p->left] = p;
                traverse.push(p->left);
            }
        }
    }

    void findPathMaxSum(TreeNode* leaf, unordered_map<TreeNode*, TreeNode*> parent) {
        TreeNode* current = leaf;
        stack<TreeNode*> stk;
        int path_max = INT_MIN;
        int path_sum = 0;

        while (current) {
            stk.push(current);
            current = parent[current];
        }

        while (!stk.empty()) {
            current = stk.top();
            stk.pop();
            path_sum += current->val;
            path_max = path_max > path_sum ? path_max : path_sum;
        }

        global_max = global_max > path_max ? global_max : path_max;
    }
};

解决了

我接受了@Dave Galvin 的建议，它奏效了！这是代码：

/**
 * 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
     */
    int maxPathSum2(TreeNode *root) {
        if (root == NULL) return 0;
        int global_max = INT_MIN;
        stack<TreeNode*> traverse;
        traverse.push(root);
        while(!traverse.empty()) {
            TreeNode* p = traverse.top();
            global_max = global_max > p->val ? global_max : p->val;
            traverse.pop();
            if (p->right) {
                traverse.push(p->right);
                p->right->val += p->val;
            }
            if (p->left) {
                traverse.push(p->left);
                p->left->val += p->val;
            }
        }
        return global_max;
    }
};
4

3 回答 3

1

我想您的代码的问题在于，当您遍历树时，在每个节点中您都在迭代以计算最大路径。最终的复杂度为O(n^2). 您需要计算流上的最大路径（在遍历树时）。

在下面的解决方案中，我使用了这里的后序迭代算法。请原谅我用这个代替了你的。

解决方案 ( ) 是简单地为每个节点O(n)添加一个字段，并在访问实际节点时取 和 之间的最大值：max_pathleftright

void postOrderTraversalIterative(BinaryTree *root) {
    if (!root) return;
    stack<BinaryTree*> s;
    s.push(root);
    BinaryTree *prev = NULL;
    while (!s.empty()) {
        BinaryTree *curr = s.top();
        if (!prev || prev->left == curr || prev->right == curr) {
            if (curr->left)
                s.push(curr->left);
            else if (curr->right)
                s.push(curr->right);
        } else if (curr->left == prev) {
            if (curr->right)
                s.push(curr->right);
        } else {
            //Visiting the node, calculate max for curr
            if (curr->left == NULL && curr->right==NULL)
                curr->max_path = curr->data;
            else if (curr->left == NULL)
                curr->max_path = curr->right->max_path + curr->data;
            else if (curr->right == NULL)
                curr->max_path = curr->left->max_path + curr->data;
            else //take max of left and right
                curr->max_path = max(curr->left->max_path, curr->right->max_path) + curr->data;
            s.pop();
        }
        prev = curr;
    }
}
于 2016-09-24T15:42:00.427 回答
0

编辑：

你不需要findPathMaxSum函数。我还更改了父地图。现在它存储 2 个值：

  1. 指向父级的指针
  2. 从根到当前节点的路径总和。

这是代码。

class Solution {
public:
    /**
     * @param root the root of binary tree.
     * @return an integer
     */
    int maxPathSum2(TreeNode *root) {
        if (root == NULL) return 0;
        findLeaf(root);
        return global_max;
    }

private:
    int global_max = INT_MIN;

    void findLeaf(TreeNode* root) {
        unordered_map<TreeNode*, std::pair<TreeNode*,int> > parent;
        stack<TreeNode*> traverse;
        parent[root] = make_pair(NULL,root->val);
        traverse.push(root);

        while(!traverse.empty()) {
            TreeNode* p = traverse.top();
            traverse.pop();
            if (!p->left && !p->right) {
                // findPathMaxSum(p, parent);
                global_max=std::max(global_max,parent[p].second);
            }
            if (p->right) {
                parent[p->right] = make_pair(p, (p->right->val) +parent[p].second) ;
                traverse.push(p->right);
            }
            if (p->left) {
                parent[p->left] = make_pair(p, (p->left->val) +parent[p].second) ;
                traverse.push(p->left);
            }
        }
    };

老的：

您想通过引用而不是findPathMaxSum中的值传递地图。

void findPathMaxSum(TreeNode* leaf, unordered_map<TreeNode*, TreeNode*> parent)

将其更改为。

void findPathMaxSum(TreeNode* leaf, unordered_map<TreeNode*, TreeNode*>& parent)

它使您的时间复杂度为 O(n*n)。

它的运行复杂度将变为 O(n*log n)。尽管由于您的约束更严格，它没有成功。所以我在上面发布了一个 O(n) 解决方案。

于 2016-09-24T14:58:50.873 回答
0

从上往下，通过添加其父节点的值来更新每个节点。跟踪您的最大值及其位置。最后返回那个。在）。

例如，如果您的二叉树是 T=[-4,2,6,-5,2,1,5]，那么我们将其更新为：[-4, 2-4=-2, 6-4=2, -2-5 = -7, -2+2=4, 2+3=3, 2+5=7]

这里的答案是 7，然后是 -4、6、5。

于 2016-09-24T15:20:40.910 回答