Leetcode 257. Binary Tree Paths

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”]

用深度优先搜索+递归处理

/** * 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:    void dfs(TreeNode* node,vector<string> & vec, string k){        if(!node->left&&!node->right){            vec.push_back(k);        }        if(node->left)            dfs(node->left,vec,k+"->"+to_string(node->left->val));        if(node->right)            dfs(node->right,vec,k+"->"+to_string(node->right->val));    }    vector<string> binaryTreePaths(TreeNode* root) {        vector<string> result;        if(NULL==root)            return result;        dfs(root,result,to_string(root->val));        return result;    }};

循环版本：

/** * 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> res;        if(NULL==root)            return res;        stack<TreeNode*> node;        stack<string> path;        node.push(root);        path.push(to_string(root->val));        while(!node.empty()){            TreeNode* curNode=node.top();            node.pop();            string curStr=path.top();            path.pop();            if(!curNode->left&&!curNode->right){                res.push_back(curStr);            }            if(curNode->left){                node.push(curNode->left);                path.push(curStr+"->"+to_string(curNode->left->val));            }            if(curNode->right){                node.push(curNode->right);                path.push(curStr+"->"+to_string(curNode->right->val));            }        }        return res;        }};