letecode [437] - Path Sum III
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.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8 10 / \ 5 -3 / \ \ 3 2 11 / \ \ 3 -2 1 Return 3. The paths that sum to 8 are: 1. 5 -> 3 2. 5 -> 2 -> 1 3. -3 -> 11
题目大意:
给定二叉树,求路径之和满足目标值的路径数量。不要求从根节点开始,也不要求到叶节点为止。即路径可以为任意一段路径,但必须从上往下。
理 解:
根据题意,在每一层它满足条件的情况都与上一层都关。对于第一层,和目标值作比较即可。第二层,需要和目标值作比较或者与目标值-第一层差值作比较。
后面的层一次类推,直接和目标值比较或者与目标值-上一层余值的差值比较。因为每一层都可能是路径起始节点。
使用vector保存上一层余值和目标值,若当前层值等于余值或目标值,则找到一个路径。同时,需要减去当前层的值更新vector中的值以便下一层计算。
代 码 C++:
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int countPath(TreeNode* root, int sum,vector<int> vec){ if(root==NULL) return 0; int count = 0; vec.push_back(sum); for(int i=0;i<vec.size();++i){ if(vec[i]==root->val) count++; vec[i] -= root->val; } count += countPath(root->left,sum,vec) + countPath(root->right,sum,vec); return count; } int pathSum(TreeNode* root, int sum) { if(root==NULL) return 0; int count = 0; vector<int> vec; count += countPath(root,sum,vec); return count; } };
运行结果:
执行用时 :72 ms, 在所有 C++ 提交中击败了8.21%的用户
内存消耗 :63.2 MB, 在所有 C++ 提交中击败了5.29%的用户