LeetCode Daily Question 30

May 17th 2022

5 月 14 日 每日一题

105. Construct Binary Tree from Preorder and Inorder Traversal

One way to serialize a binary tree is to use preorder 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 ‘#’.

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 preorder, return true if it is a correct preorder traversal serialization of a binary tree.

It is guaranteed that 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”.
Note: You are not allowed to reconstruct the tree.

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Example 1:

Input: preorder = "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true
Example 2:

Input: preorder = "1,#"
Output: false
Example 3:

Input: preorder = "9,#,#,1"
Output: false

思路

本题有两种写法，我用了最笨的一种，每遇到两个#，将这两个 hash 和对应的 leaf node 删掉，然后添加一个#，要注意的是需要使用 while 循环，因为可能会在添加#之后形成新的满足 pop 要求的组合。要注意的是不能直接用 index 访问，因为可能会出现两位数字（92）。
另一种写法是聪明很多，因为 external node 数量必然比 internal 多 1，每遇到一个 val，counter++，遇到#，counter–，如果读完所有的 node 时，count 为-1，则有效（如果未读完所有 node，count 就为 1，意味着此时已经无法插入 node，则此 tree 无效）。
截屏2022-05-18 下午8.44.49

代码

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isValidSerialization(string preorder) {
        //if it's #
        if(preorder[0] == '#' && preorder.size() == 1)
            return  true;
        else if(preorder[0] == '#' && preorder.size() != 1)
            return false;
        preorder += ',';
        stringstream str(preorder);
        string ele;
        getline(str, ele, ',');
        vector<string> stack;
        stack.push_back(ele);
        while(getline(str, ele, ',')){
            stack.push_back(ele);
            if(stack.back() == "#"){
                while(stack.size() > 2 && stack.back() == "#" && stack[stack.size() - 2] == "#" && stack[stack.size() - 3] != "#" ){
                    stack.pop_back();
                    stack.pop_back();
                    stack.pop_back();
                    stack.push_back("#");
                    }
                }
        }
        if(stack.size() == 1 && stack.back() == "#")
            return true;
        return false;
    }
};

bool isValidSerialization(string p)
{
    stringstream ss{p};
	int c{0};
	for( ; c!=-1 and !ss.eof(); getline(ss, p, ','))
        p[0]=='#' ? --c : ++c；
	return c==-1 and ss.eof();
}
,
  1. 5 月 14 日 每日一题
    1. 105. Construct Binary Tree from Preorder and Inorder Traversal
    2. 思路
    3. 代码