606 Construct String from Binary Tree

606. Construct String from Binary Tree

1. Question

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:
1
Input:
2
Binary tree: [1,2,3,4]
3
1
4
/ \
5
2 3
6
/
7
4
8
9
10
Output:
11
"1(2(4))(3)"
12
13
14
15
Explanation:
16
Originallay it needs to be "1(2(4)())(3()())",
17
18
19
but you need to omit all the unnecessary empty parenthesis pairs.
20
21
22
And it will be "1(2(4))(3)".
Copied!
Example 2:
1
Input:
2
Binary tree: [1,2,3,null,4]
3
1
4
/ \
5
2 3
6
\
7
4
8
9
10
Output:
11
"1(2()(4))(3)"
12
13
14
15
Explanation:
16
Almost the same as the first example,
17
18
19
except we can't omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.
Copied!

2. Implementation

(1) Recursion
思路:直接preorder traversal就好,注意如果右子树是空的时候,不需要加"()",这样并不影响1-1对应的关系
1
class Solution {
2
public String tree2str(TreeNode t) {
3
if (t == null) {
4
return "";
5
}
6
7
String res = t.val + "";
8
9
String left = tree2str(t.left);
10
String right = tree2str(t.right);
11
12
if (left == "" && right == "") {
13
return res;
14
}
15
else if (left == "") {
16
return res + "()" + "(" + right + ")";
17
}
18
else if (right == "") {
19
return res + "(" + left + ")";
20
}
21
else {
22
return res + "(" + left + ")" + "(" + right + ")";
23
}
24
}
25
}
Copied!

3. Time & Space Complexity

Recursion: 时间复杂度O(n), 空间复杂度O(h)