题目:
给定一个二叉树,返回所有从根节点到叶子节点的路径。
说明: 叶子节点是指没有子节点的节点。
示例:
输入: 1 / 2 3 5 输出: ["1->2->5", "1->3"] 解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3
解题思路:
递归,在参数列表里回溯的方法灰常好用,这里介绍两种方法。
代码:
法一:
/** * 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: vector<string> binaryTreePaths(TreeNode* root) { vector<string> ans; if(root == NULL) return ans; if(!root->left && !root->right) ans.push_back(to_string(root->val)); vector<string> leftSub = binaryTreePaths(root->left); for(int i=0; i<leftSub.size(); ++i) ans.push_back(to_string(root->val) + "->" + leftSub[i]); vector<string> rightSub = binaryTreePaths(root->right); for(int i=0; i<rightSub.size(); ++i) ans.push_back(to_string(root->val) + "->" + rightSub[i]); return ans; } };
参考来源:https://blog.csdn.net/my_clear_mind/article/details/82283939
法二:
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 void DFS(TreeNode* root, string temp, vector<string> &ans) 13 { 14 if(root == NULL) { 15 return ; 16 } 17 if(root->left == NULL && root->right == NULL) { 18 temp += to_string(root->val); 19 ans.push_back(temp); 20 return ; 21 } 22 if(root->left) { 23 DFS(root->left, temp + to_string(root->val) + "->", ans); 24 } 25 if(root->right) { 26 DFS(root->right, temp + to_string(root->val) + "->", ans); 27 } 28 } 29 vector<string> binaryTreePaths(TreeNode* root) { 30 vector<string> ans; 31 if(root == NULL) 32 return ans; 33 DFS(root, "", ans); 34 return ans; 35 } 36 };