257.leetcode Binary Tree Paths(easy)[二叉树深度路径遍历]

来源：互联网发布：nba直播数据统计编辑：程序博客网时间：2024/05/18 01:34

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"]

具体思路：要深度优先遍历二叉树，那么一般都用到栈，这里用函数的栈递归来完成

/** * 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 getResult(vector<string> &result,TreeNode* root,string &s)    {        stringstream ss;        ss<<root->val;        string temp = s;        temp += ss.str();        string s1("->");        int flag1 = 0,flag2 = 0;        if(root->left != NULL)        {            temp += s1;            flag1 = 1;            getResult(result,root->left,temp);        }        if(root->right != NULL)        {            if(!flag1)               temp += s1;            flag2 = 1;            getResult(result,root->right,temp);        }        if(!flag1&&!flag2)//左右子树都为空，即叶子节点的时候一条路径完成        {            result.push_back(temp);        }    }    vector<string> binaryTreePaths(TreeNode* root) {        vector<string> result;        if(root == NULL)            return result;        string s;        getResult(result,root,s);        return result;    }};

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