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.
根据给定的二叉树返回根据其建立的字符串，建立原则按照先序序列顺序，有子节点的用括号括起来，叶子节点的孩子括号省略。特别的，若左孩子不存在而有孩子存在，则左孩子的括号不可省略。
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
递归调用即可。
/**
* 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:
string tree2str(TreeNode* t) {
if(t==nullptr) return "";
string res;
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;
}
};
