LintCode 1137: Construct String from Binary Tree (经典二叉树题目)

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1137. Construct String from Binary Tree
      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.

思路：二叉树递归。
记得左子树没有右子树有的情况仍然要输出()，这是题目的要求。
代码如下：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param t: the root of tree
     * @return: return a string
     */
    string tree2str(TreeNode * t) {
        string result;
        helper(t, result);
        return result;
    }
    
    void helper(TreeNode * node, string & s) {
        if (node) {
            string s1;
            helper(node->left, s1);
            string s2;
            helper(node->right, s2);
            
            if ((s1.size() == 0) && (s2.size() == 0))
                s += to_string(node->val);
            else if ((s1.size() != 0) && (s2.size() == 0))
                s += to_string(node->val) + '(' + s1 + ')';
           // else if ((s1.size() == 0) && (s2.size() != 0))
           //   s += to_string(node->val) + '(' + s2 + ')';
            else    
                s += to_string(node->val) + '(' + s1 + ')' + '(' + s2 + ')';
        } 
    }
};