LeetCode 437 Path Sum III

本文轉載自查看原文 2016-11-07 23:35 2649 LeetCode Solutions - Easy

Problem:

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Summary:

找到二叉樹中的子路徑條數使得該子路徑權值之和與給定和相等。

Analysis:

本題參考：http://www.cnblogs.com/grandyang/p/6007336.html

1. 最直觀的思路：通過前序遍歷的方式遍歷二叉樹，每遍歷至一個節點，將當前節點的權值記錄在數組中，並更新記錄當前路徑和的變量curSum，同時判斷更新后的curSum與 sum是否相等，若相等則res加1。

此時的curSum為從根節點到當前節點的路徑之和。至於這條路徑上的子路徑是否滿足題意，則需將數組中的權值在curSum中一個個減掉，每剪掉一個，判斷一次當前權值和是否與sum相等，若相等則res加1。此處需注意不能將數組中的權值完全減掉，最后應保留一個，因為全部去掉所得curSum為0，若題中所給的sum剛好為0，則判斷相等，但不符合題意。

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int pathSum(TreeNode* root, int sum) {
13         int res = 0;
14         vector<TreeNode*> vec;
15         findPath(root, sum, 0, res, vec);
16         
17         return res;
18     }
19     
20     void findPath(TreeNode* node, int sum, int curSum, int &res, vector<TreeNode*> &vec) {
21         if (!node) return;
22         curSum += node->val;
23         vec.push_back(node);
24         
25         if (sum == curSum) res++;
26         int tmp = curSum;
27         for (int i = 0; i < vec.size() - 1; i++) {
28             tmp -= vec[i]->val;
29             if (tmp == sum) res++;
30         }
31         
32         findPath(node->left, sum, curSum, res, vec);
33         findPath(node->right, sum, curSum, res, vec);
34         
35         vec.pop_back();
36     }
37 };

2. 優化算法，用hash表記錄當前遍歷路徑中的子路徑權值和對應出現的次數。

若sum為從根節點到某x節點的路徑權值和，則遍歷至節點x時，當前的路徑和curSum恰好與sum相等，則res = m[curSum - sum] = m[0] = 1;
若sum為某段子路徑權值和，如：x₁->x₂->x₃->x₄......中sum等於節點x₃與節點x₄的權值和，即sum = sum_x3+x4。則遍歷至x₂時， m[curSum]++; 處已經記錄了m[curSum] = m[sum_x1+x2] = 1,便利至x₄時curSum = sum_x1+x2+x3+x4,則res = m[curSum - sum] = m[sum_x1+x2+x3+x4 - sum_x3+x4] = m[sum_x1+x2] = 1。

 1 class Solution {
 2 public:
 3     int pathSum(TreeNode* root, int sum) {
 4         unordered_map<int, int> m;
 5         m[0] = 1; 
 6         int res = findPath(root, sum, 0, m);
 7         
 8         return res;
 9     }
10     
11     int findPath(TreeNode* node, int sum, int curSum, unordered_map<int, int> &m) {
12         if (!node) return 0;
13         curSum += node->val;
14         
15         int res = m[curSum - sum];
16         m[curSum]++;
17         res += (findPath(node->left, sum, curSum, m) + findPath(node->right, sum, curSum, m));
18         m[curSum]--;
19         
20         return res;
21     }
22 };

3. 最簡潔的方法，以每一個節點作為路徑根節點進行前序遍歷，查找每一條路徑的權值和與sum是否相等。

 1 class Solution {
 2 public:
 3     int pathSum(TreeNode* root, int sum) {
 4         if (!root) return 0;
 5         int res = findPath(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
 6         return res;
 7     }
 8     
 9     int findPath(TreeNode* node, int curSum, int sum) {
10         if (!node) return 0;
11         curSum += node->val;
12         return (curSum == sum) + findPath(node->left, curSum, sum) + findPath(node->right, curSum, sum);
13     }
14 };

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