Leetcode Serialize and Deserialize Binary Tree

The original title link is here: https://leetcode.com/problems/serialize-and-deserialize-binary-tree/

Put a tree serialize and deserialize. The recursive method is serialize by pre-order.

    1   /   2   3     /     4   5

The example becomes the "1,2,#,#,3,4,#,#,5,#,#". Then deserialize reads a character as the root node, and then the left and right child nodes of the root node call the Deserializehelper function recursively.

Here Stringtokenizer:stringtokenizer st = new StringTokenizer (data, ",") can be used to separate data with "," to generate StringTokenizer.

StringTokenizer two Method:st.hasMoreTokens () returns whether tokens and St.nexttoken () return the next string. StringTokenizer is much like String.Split ().

But he can process the split string in the middle.

AC Java:

1  Public classCodec {2 3     //encodes a tree to a single string.4      PublicString serialize (TreeNode root) {5StringBuilder SB =NewStringBuilder ();6 Serializehelper (ROOT,SB);7         returnsb.tostring ();8     }9     Ten     Private voidserializehelper (TreeNode root, StringBuilder sb) { One         if(Root = =NULL){ ASb.append ("#,"); -             return; -         } theSb.append (Root.val). Append (","); - Serializehelper (Root.left, SB); - Serializehelper (Root.right, SB); -     } +  -     //decodes your encoded data to tree. +      PublicTreeNode Deserialize (String data) { AStringTokenizer st =NewStringTokenizer (data, ","); at         returnDeserializehelper (ST); -     } -      -     PrivateTreeNode Deserializehelper (StringTokenizer St) { -         if(!St.hasmoretokens ()) { -             return NULL; in         } -String cur =st.nexttoken (); to         if(Cur.equals ("#")){ +             return NULL; -         } theTreeNode root =NewTreeNode (integer.valueof (cur)); *Root.left =Deserializehelper (ST); $Root.right =Deserializehelper (ST);Panax Notoginseng         returnRoot; -     } the}

The iteration method uses the BFS. When processing serialize, queue is not empty, popup node cur, if cur is null, add "#," after SB; If it is not NULL, add Cur.val and "," after SB.

When processing deserialize, first string array with "," Gecheng string. If it is full tree, then the left node index is 2*i+1, the right node index is 2*i+2. But this is not full of trees.

It is necessary to record at the same time that there are several null nodes in front of the current I, with Count storage, the left node index into a (i-count[i]) + 1, the right node index into a (I-count[i]) + 2.

Note: Remember that the queue is an abstract class and cannot be used directly to generate que, it needs to be linkedlist.

AC Java:

1  Public classCodec {2 3     //encodes a tree to a single string.4      PublicString serialize (TreeNode root) {5StringBuilder SB =NewStringBuilder ();6Queue<treenode> que =NewLinkedlist<treenode> ();//Error7 Que.offer (root);8          while(!Que.isempty ()) {9TreeNode cur =Que.poll ();Ten             if(cur = =NULL){ OneSb.append ("#,"); A}Else{ -Sb.append (Cur.val). Append (","); - Que.offer (cur.left); the Que.offer (cur.right); -             } -         } -         returnsb.tostring (); +     } -  +     //decodes your encoded data to tree. A      PublicTreeNode Deserialize (String data) { atString [] s = Data.split (","); -         int[] Count =New int[s.length]; -TreeNode [] TNS =NewTreenode[s.length]; -          -          for(inti = 0; i<s.length; i++){ -             if(i>0){ inCount[i] = count[i-1]; -             } to             if(S[i].equals ("#")){ +Tns[i] =NULL; -count[i]++; the}Else{ *Tns[i] =NewTreeNode (integer.valueof (s[i)); $             }Panax Notoginseng         } -          the          for(inti = 0; i<s.length; i++){ +             if(S[i].equals ("#")){ A                 Continue; the             } +Tns[i].left = tns[2 * (i-count[i]) + 1]; -Tns[i].right = tns[2 * (I-count[i]) + 2]; $         } $         returnTns[0]; -     } -}

Reference this post: http://blog.csdn.net/ljiabin/article/details/49474445

Leetcode Serialize and Deserialize Binary Tree