LeetCode:Verify Preorder Serialization of a Binary Tree

Posted 2020-06-17

One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node‘s value. If it is a null node, we record using a sentinel value such as #.

     _9_
    /      3     2
  / \   /  4   1  #  6
/ \ / \   / # # # #   # #

For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.

Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

Each comma separated value in the string must be either an integer or a character ‘#‘ representing null pointer.

You may assume that the input format is always valid, for example it could never contain two consecutive commas such as "1,,3".

 

题目：判断二叉树的先序序列正确性（不重建树）

思路：二叉树先序遍历序列类似于进栈出栈的过程，可以理解为每个新节点的产生都新增需要一个 ‘#‘ 在序列中。根节点需要两个 ‘#‘ ，因此节点总数应该比 ‘#‘ 总数少一个。

做法：借助一个栈，遇数进栈，遇 ‘#‘ 出栈，遍历到序列最后应该是多出一个 ‘#‘ 。另外特殊考虑空树的情况即可。

 

 1 class Solution {
 2 public:
 3     bool isValidSerialization(string preorder) {
 4         stack<string> stack1;
 5         string tmpstr = "";
 6         int i = 0;
 7         int length = preorder.size();
 8         bool getnumber = 0;
 9         if(length == 1 && preorder[0] == ‘#‘){
10             return true;
11         }
12         while(i < length - 1){
13             if(preorder[i] == ‘#‘){
14                 if(stack1.empty()){
15                     return false;
16                 }else{
17                     stack1.pop();
18                     i ++;
19                 }
20             }else if(preorder[i] == ‘,‘){
21                 if(getnumber){
22                     getnumber = false;
23                     stack1.push(tmpstr);
24                     tmpstr = "";
25                     i++;
26                 }else{
27                     i++;
28                 }
29             }else{
30                 getnumber = true;
31                 tmpstr += string(1, preorder[i]);
32                 i++;
33             }
34         }
35         if(stack1.empty() && preorder[i] == ‘#‘){
36             return true;
37         }
38         else{
39             return false;
40         }
41     }
42 };