leetcode257.二叉树的所有路径

解法1：DFS

这题的思路是使用前序遍历，然后记住string参数这里只能值传递，不能引用传递，只有使用值传递才能保证每个调用栈才能使用不同的string副本！

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> v;
        string s;
        dfs(root,v,s);
        return v;
    }
    
    void dfs(TreeNode* node,vector<string>& v,string s){
        if(node){
            s.append(to_string(node->val));
            if(!node->left&&!node->right){
                v.push_back(s);
                
            }else{
                s.append("->");
                dfs(node->left,v,s);
                dfs(node->right,v,s);
            }
        }
    }
};
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解法2：BFS

手写一算就懂了！

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> v;
        queue<TreeNode*> node_q;
        queue<string> path_q;
        bfs(v,root,node_q,path_q);
        return v;
    }
    void bfs(vector<string>& v,TreeNode* root,queue<TreeNode*>& node_q,queue<string>& path_q){
        node_q.push(root);
        path_q.push(to_string(root->val));
        while(!node_q.empty()){
            TreeNode* node=node_q.front();
            string path = path_q.front();
            node_q.pop();
            path_q.pop();
            if(!node->left&&!node->right){
                v.push_back(path);
            }else{
                if(node->left){
                    node_q.push(node->left);
                    path_q.push(path+"->"+to_string(node->left->val));
                }
                if(node->right){
                    node_q.push(node->right);
                    path_q.push(path+"->"+to_string(node->right->val));
                }
            }
        }
    }
};
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