我正在努力解決這個問題,我想以非遞歸方式解決這個問題。在我的算法中似乎沒有邏輯錯誤,通過了73%的測試用例。但它無法處理大數據,報告「超出時間限制」。我很感激,如果有人可以給我一些提示如何在非遞歸的情況下做到這一點,並避免超時,請提前致謝!二叉樹最大路徑總和,非遞歸,超過時間限制
問題鏈接
我相信也有一個類似的本文給出了。
http://www.lintcode.com/en/problem/binary-tree-maximum-path-sum-ii/
問題描述:
給定一個二叉樹,找到根的最大路徑總和。 該路徑可能在樹中的任何節點處結束,並且至少包含一個節點。
實施例:
鑑於下面二叉樹:
/\
回報4.(1-> 3)
法官
時間超出限制
總運行時間:1030毫秒
輸入 輸入數據
{-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,#,#, - 333754,#817842,-220,-269,9,-862,-78,-473,643,536 ,-142,773,485,262,360,702,-661,244,-96,#519566,-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,# ,202213,-141,759,214,207,135,329,15,#,#,244#,334628509627,-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,-601,43 5,-114,-337,422,#, - 855,-859,163,-224,902,#,577,#, - 386,272,-9 ...
預計
我的代碼 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;
}
};
解決
我接受@Dave Galvin的建議,它的工作原理!下面的代碼:
/**
* 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;
}
};
爲什麼它是必須非遞歸? – Tempux
@ sudomakeinstall2我只是想以非遞歸的方式嘗試它.... – xman
你能提供鏈接到問題的來源嗎? –