題目:
Given a binary tree, return all root-to-leaf paths.
For example, given the following binary tree:
1 / \ 2 3 \ 5
All root-to-leaf paths are:
["1->2->5", "1->3"]就是一個深搜,關鍵點是,什麼時候把搜到的加上去。它這個是每個葉子都有一條深搜的道路。也就是在遞歸的時候就將結點加上去了,於是就有了所有的道路。當到葉子的時候把這個當前的道路加到要輸出的總和中,然後繼續下面的遞歸。
然後要注意的是他給的root可能是空的,要先判斷一下。
還有就是int要轉成string,中間還要加->
貼代碼:
/**
* 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> outstring;
if(root==NULL) return outstring;
search(outstring,root,to_string(root->val));
return outstring;
}
void search(vector<string>&outstring, TreeNode* root, string value){
if(root->left==NULL && root->right == NULL){
outstring.push_back(value);
}
if(root->left != NULL){
search(outstring, root->left, value + "->" + to_string(root->left->val));
}
if(root->right != NULL){
search(outstring, root->right, value + "->" + to_string(root->right->val));
}
return;
}
};