leetcode 257. Binary Tree Paths

来源：互联网发布：linux查找文件夹编辑：程序博客网时间：2024/05/17 05:55

/*leetcode 257. Binary Tree PathsGiven a binary tree, return all root-to-leaf paths.For example, given the following binary tree:  1/   \2    3\ 5All root-to-leaf paths are:["1->2->5", "1->3"]题目大意：求解二叉树中所有根节点到叶子节点的路径解题思路：*/#include "TreeInclude.h"#include <unordered_set>#include <stack>class Solution{public:    //iterator    vector<string> binaryTreePaths(TreeNode* root)    {        vector<string> ret;        if (!root)            return ret;        //存放路径的栈        stack<TreeNode*> stk;        stk.push(root);        //存放路径的数        vector<int> pathVal({root->val});        //是否已经遍历过了        unordered_set<TreeNode*> umap;        umap.insert(root);        TreeNode* cur;        while (stk.size())        {            cur = stk.top();            //如果左右子树都为空则：            //找到一条路径，压入返回数组中；继续回溯            if (!cur->left && !cur->right)            {                ret.push_back(GenHelper(pathVal));                pathVal.pop_back();                stk.pop();            }            else             {                if (cur->left)                 {                    auto isLeft = umap.find(cur->left);                    if (isLeft == umap.end())//没有访问过                    {                        //开始访问：把值存放在值得数组中，把节点放到栈中，并标记为已访问                        pathVal.push_back(cur->left->val);                        stk.push(cur->left);                        umap.insert(cur->left);                        continue;                    }                }                //能够到这里：说明左子树为空或者左子树已经访问过了                //因为我们要每次访问一条路径，因此不能把两条路径混杂了。                if (cur->right)                {                    auto isRight = umap.find(cur->right);                    if (isRight == umap.end())//没有访问过                    {                        //开始访问：把值存放在值得数组中，把节点放到栈中，并标记为已访问                        pathVal.push_back(cur->right->val);                        stk.push(cur->right);                        umap.insert(cur->right);                        continue;                    }                }                //左右子树都访问完了                pathVal.pop_back();                stk.pop();            }        }        return ret;    }    string GenHelper(vector<int> v)    {        int len = v.size();        string ret;        if (len == 0)            return ret;        ret += to_string(v[0]);        for (int i = 1; i < len; ++i)            ret += ("->" + to_string(v[i]));        return ret;    }};class Solution1{public:    //recusive    vector<string> binaryTreePaths(TreeNode* root)    {        vector<string> ret;        if (!root)            return ret;        FindHelp(ret, root, to_string(root->val));        return ret;    }    void FindHelp(vector<string>& ret, TreeNode* root, string str)    {        //退出条件：左右子树都为空        if (!root->left && !root->right)        {            ret.push_back(str);            return;        }        if (root->left)        {            FindHelp(ret, root->left, str + "->" + to_string(root->left->val));        }        if (root->right)        {            FindHelp(ret, root->right, str + "->" + to_string(root->right->val));        }    }};void TEST(){    Solution sol;    vector<int> v{ 1,2,3,-1,5 };    TreeNode* root = CreateTree(v);    vector<string> ret = sol.binaryTreePaths(root);    for (auto str : ret)        cout << str << endl;}int main(){    TEST();    return 0;}

头文件

#ifndef _TREE_INCLUDE_H_ #define _TREE_INCLUDE_H_#include <iostream>#include <vector>#include <algorithm>#include <string>using namespace std;// 节点数据结构struct TreeNode{    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};//使用数组创建二叉树void CreateNode(const vector<int>& v, TreeNode* cur, int curIndex){    int len = v.size();    if (len == 0 || cur == NULL || curIndex < 0 || curIndex >= len)        return;    if ((2 * curIndex + 1 < len) && (v[2*curIndex+1] != -1))    {        TreeNode* tmp = new TreeNode(v[2 * curIndex + 1]);        cur->left = tmp;    }    else        cur->left = NULL;    if ((2 * curIndex + 2 < len) && (v[2*curIndex+2] != -1))    {        TreeNode* tmp = new TreeNode(v[2 * curIndex + 2]);        cur->right = tmp;    }    else        cur->right = NULL;    CreateNode(v, cur->left, 2 * curIndex + 1);    CreateNode(v, cur->right, 2 * curIndex + 2);}TreeNode* CreateTree(const vector<int>& v){    int len = v.size();    if (len == 0)        return NULL;    TreeNode* root = new TreeNode(v[0]);    CreateNode(v, root, 0);    return root;}void FirstOrderTraverse(TreeNode* root){    if (root == NULL)        return;    cout << root->val << " ";    FirstOrderTraverse(root->left);    FirstOrderTraverse(root->right);    //cout << endl;}void MakeEmpty(TreeNode* root){    if (root != NULL)    {        MakeEmpty(root->left);        MakeEmpty(root->right);        delete root;    }}#endif //

0 0