【题目】
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] ]
【题意】
判断二叉树中是否存在一条从根到叶子节点的路径,使得路径上的节点值之和等于所给的值,输出所有的路径
【思路】
DFS,递归
【代码】
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void dfs(TreeNode*root, vector<vector<int> >&result, vector<int>path, int pathSum, int sum){ path.push_back(root->val); pathSum+=root->val; if(root->left==NULL && root->right==NULL){ if(pathSum==sum) result.push_back(path); return; } //搜索左子树 if(root->left) dfs(root->left, result, path, pathSum, sum); //搜索右子树 if(root->right) dfs(root->right, result, path, pathSum, sum); } vector<vector<int> > pathSum(TreeNode *root, int sum) { vector<vector<int> >result; if(root==NULL)return result; vector<int>path; int pathSum=0; dfs(root, result, path, pathSum, sum); return result; } };
LeetCode: Path Sum II [113],布布扣,bubuko.com
时间: 2024-10-12 08:02:25