Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.
For example:
Given the below binary tree and sum
,
= 22
5 / 4 8 / / 11 13 4 / \ / 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
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class Solution { public: vector<vector<int>> pathSum(TreeNode* root, int sum) { vector<vector<int>> res; vector<int> tmp; //保存中间结果 tmpFunction(root, sum, tmp, res); return res; } void tmpFunction(TreeNode* root, int sum, vector<int> &tmp, vector<vector<int>>&res){ if (root == NULL) return; tmp.push_back(root->val); if (root->left == NULL && root->right == NULL){ if (root->val == sum) res.push_back(tmp); } tmpFunction(root->left, sum - root->val, tmp, res); tmpFunction(root->right, sum - root->val, tmp, res); tmp.pop_back(); } };
#include<iostream> #include<new> #include<vector> using namespace std; //Definition for binary tree struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { public: vector<vector<int> > pathSum(TreeNode *root, int sum) { vector<vector<int> > path; vector<int> tmp; hasPathSum(root, sum, path, tmp); return path; } void hasPathSum(TreeNode *root, int sum, vector<vector<int> > &path, vector<int> tmp) { if (root == NULL) return; tmp.push_back(root->val); if (root->left == NULL&&root->right == NULL && (sum - root->val) == 0) { path.push_back(tmp); } if (root->left) hasPathSum(root->left, sum - root->val, path, tmp); if (root->right) hasPathSum(root->right, sum - root->val, path, tmp); tmp.pop_back(); } void createTree(TreeNode *&root) { int i; cin >> i; if (i != 0) { root = new TreeNode(i); if (root == NULL) return; createTree(root->left); createTree(root->right); } } }; int main() { Solution s; TreeNode *root; s.createTree(root); vector<vector<int> > path = s.pathSum(root, 22); for (auto a : path) { for (auto v : a) cout << v << " "; cout << endl; } system("pause"); return 0; }
时间: 2024-10-12 04:47:21