LeetCode 606. Construct String from Binary Tree

606. Construct String from Binary Tree

Description

You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way.
The null node needs to be represented by empty parenthesis pair “()”. And you need to omit all the empty parenthesis pairs that don’t affect the one-to-one mapping relationship between the string and the original binary tree.

Example 1:Input: Binary tree: [1,2,3,4]       1     /   \    2     3   /      4     Output: "1(2(4))(3)"Explanation: Originallay it needs to be "1(2(4)())(3()())", but you need to omit all the unnecessary empty parenthesis pairs. And it will be "1(2(4))(3)".Example 2:Input: Binary tree: [1,2,3,null,4]       1     /   \    2     3     \        4 Output: "1(2()(4))(3)"Explanation: Almost the same as the first example, except we can't omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

Solution

• 题目意思是将一棵二叉树转换成字符串表示，样例在上面，就是一种类似广义表的形式。
• 我的思路是利用递归，先保存根值，在递归进行子树的字符串转换，值得注意的是处理不同情况下的括号添加，我的代码如下：

char s[100000];void travel(struct TreeNode *t) {    if (!t) return ;    char temp[100];    sprintf(temp,"%d",t->val);    strcat(s,temp);    if (!t->left && !t->right) return ;   //如果左右子树都没有，此时不应该保存括号进去    strcat(s,"(");    travel(t->left);    strcat(s,")");    if (t->left && !t->right) return ;    //如果只有左子树没有右子树，右子树对应的括号也不应该保存进去，但值得注意的是，如果左子树没有但是右子树有，此时左子树的括号还是应当保存进去    strcat(s,"(");    travel(t->right);    strcat(s,")");}char* tree2str(struct TreeNode* t) {    memset(s,0,sizeof(s));    travel(t);    return s;}