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. use recursive way
e.g.
1
/ 2 3
/
4 5
2's subtree have the similar substructure, having the similar problem. 2(4)(5)
inside the whole answer 1(2(4)(5))(3)
so it is easy to use recursive way to construct the string
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
2. use iterative way
how to add right parentheses ")"? one way is we can judge to visit the node again to complete the ")". Hence if it is visited before, then we add ")", also pop from the stack
In one word, we use visited node set to record visited or not, and also stack to record traverse nodes as normal DFS
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