c++ - 二叉树最大路径和，非递归，超过时间限制

Question

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。但是无法处理大数据，报“Time Limit Exceeded”。如果有人能给我一些提示，说明如何以非递归方式执行此操作并避免超过时间限制，我将不胜感激，在此先感谢!

问题链接

I believe there's also a similar one in LeetCode.

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

问题描述:

Given a binary tree, find the maximum path sum from root. The path may end at any node in the tree and contain at least one node in it.

示例:

Given the below binary tree:

1

/ \

2 3

返回 4.(1->3)

判断

Time Limit Exceeded

Total Runtime: 1030 ms

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,-705,868,-782,300,-945,-3,139,-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,-215,-555,132,-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,#,-855,-859,163,-224,902,#,577,#,-386,272,-9 ...

预期

6678

我的代码C++语言

/**
 * 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;
    }
};

已解决

我接受@Dave Galvin 的建议并且它有效!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;
    }
};

Answer 1

我猜你的代码的问题在于，当你遍历你的树时，你在每个节点中迭代以计算最大路径。最终复杂度为 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 here 中的后序迭代算法.请原谅我用这个代替你的。

解决方案(O(n))就是简单地为每个节点添加一个字段max_path，当访问时实际节点取left 和 right 之间的最大值:

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;
    }
}