[Leetcode] Binary tree-- 606. Construct String from Binary Tree

标签：substr   -o   int   pairs   break   present   remove   xpl   end   

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: 

1         if t is None:
2             return ""
3         strRes = str(t.val)
4         
5         if t.left or t.right:
6             strRes += "(" + self.tree2str(t.left) + ")"
7         if  t.right:
8             strRes += "(" + self.tree2str(t.right) + ")"
9         return strRes

 1         stk = []           #stack
 2         visitedSet = set()
 3         strRes = ""
 4         if t is not None:
 5             #put t into stack
 6             stk.append(t)
 7         while (len(stk)):
 8             node = stk[-1]
 9             #print ("node: ", node.val)
10             if node not in visitedSet:
11                 visitedSet.add(node)
12                 #add "(" + val
13                 strRes += "(" + str(node.val)
14                 if node.right is not None:
15                     stk.append(node.right)
16                 if node.left is not None:
17                     stk.append(node.left)
18                 elif node.right is not None:
19                     strRes += "()"                 #make sure it doesn‘t affect the one-to-one mapping relationship
20                     
21             else:
22                 strRes += ")"
23                 #pop node
24                 stk.pop()
25         return strRes[1:-1]                   #remove the first and last parenthesis

[Leetcode] Binary tree-- 606. Construct String from Binary Tree

标签：substr   -o   int   pairs   break   present   remove   xpl   end   

原文地址：http://www.cnblogs.com/anxin6699/p/7220468.html