leetcode -day29 Binary Tree Inorder Traversal & Restore IP Addresses

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1、

Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

分析：求二叉树的中序遍历，采用递归的方法的话非常简单，如果非递归的话，就需要用栈来保存上层结点，开始向左走一直走到最左叶子结点，然后将此值输出，从队列中弹出，如果右子树不为空则压入该弹出结点的右孩子，再重复上面往左走的步骤直到栈为空即可。

class Solution {public:    vector<int> inorderTraversal(TreeNode *root) {        vector<int> result;        if(!root){            return result;        }        TreeNode* tempNode = root;        stack<TreeNode*> nodeStack;        while(tempNode){            nodeStack.push(tempNode);            tempNode = tempNode->left;        }        while(!nodeStack.empty()){            tempNode = nodeStack.top();            nodeStack.pop();            result.push_back(tempNode->val);            if(tempNode->right){                nodeStack.push(tempNode->right);                tempNode = tempNode->right;                while(tempNode->left){                    nodeStack.push(tempNode->left);                    tempNode = tempNode->left;                }            }        }        return result;    }};

2、Restore IP Addresses

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

For example:
Given "25525511135",

return ["255.255.11.135", "255.255.111.35"]. (Order does not matter)

分析：此题跟我之前遇到的一个判断字符串是否是ip地址有点类似，http://blog.csdn.net/kuaile123/article/details/21600189，采用动态规划的方法，参数num表示字符串表示为第几段，如果num==4则表示最后一段，直接判断字符串是否有效，并保存结果即可，如果不是则点依次加在第0个、第1个....后面，继续递归判断后面的串。

如下：

class Solution {public:    vector<string> restoreIpAddresses(string s) {        vector<string> result;        int len = s.length();        if(len < 4 || len > 12){            return result;        }        dfs(s,1,"",result);        return result;    }    void dfs(string s, int num, string ip, vector<string>& result){        int len = s.length();        if(num == 4 && isValidNumber(s)){            ip += s;            result.push_back(ip);            return;        }else if(num <= 3 && num >= 1){            for(int i=0; i<len-4+num && i<3; ++i){                string sub = s.substr(0,i+1);                if(isValidNumber(sub)){                    dfs(s.substr(i+1),num+1,ip+sub+".",result);                }            }        }    }    bool isValidNumber(string s){        int len = s.length();        int num = 0;        for(int i=0; i<len; ++i){            if(s[i] >= '0' && s[i] <= '9'){                num = num*10 +s[i]-'0';            }else{                return false;            }        }        if(num>255){            return false;        }else{            //非零串首位不为0的判断            int size = 1;            while(num = num/10){                ++size;            }            if(size == len){                 return true;            }else{                return false;            }        }    }};

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