<easy>LeetCode Problem -- 606. Construct String from Binary Tree

描述：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.

分析：通过题目所给的二叉树来构建字符串，要求对于空节点构建出”()”,非空节点为(节点元素)，并且要在不破环一一映射的前提下删除多余的括号。

思路一：递归。典型递归题目。需要注意的是当一个节点的左子节点不存在时，如果右子节点存在，则应该用()代表右子节点，此时的这个()不能被省略。

class Solution {
public:
string tree2str(TreeNode* t) {
if (t == NULL) return "";
if (t -> right == NULL && t -> left == NULL) return to_string(t -> val);
string res = to_string(t -> val);
if (t->left) res += '(' + tree2str(t->left) + ')';
else if (t->right) res += "()";
if (t->right) res += '(' + tree2str(t->right) + ')';
return res;
}
};