[Leetcode] Construct String from Binary Tree(Easy)
LeetCode 606 - Construct String from Binary Tree
Given the root of a binary tree, construct a string consisting of parenthesis and integers from a binary tree with the preorder traversal way, and return it.
Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree. example
Input: root = [1,2,3,4]
Output: "1(2(4))(3)"
Explanation: Originally, 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)"
Input: root = [1,2,3,null,4]
Output: "1(2()(4))(3)"
Explanation: Almost the same as the first example, except we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output
How can we solve this problem?
這題就是要讓我們講以string的方式輸出Binary Tree。只要注意他的規則就可以解決這題。
- node的children都會被
()包裹住 - node如果有
left-child沒有right-child可以無視right-child的() - node如果有
right-child沒有left-child,left-child的位置必須包含一個()
根據以上這幾條輸出規則，透過postorder traversal就可以解決。
postorder traversal : traverse Left-child -> Right-child -> self node
Solution:
/**
* 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:
string tree2str(TreeNode* root) {
string res;
solution(root,res);
return res;
}
//Solution A
string constructStr(TreeNode* root){
if(!root) return "";
string cur = to_string(root->val);
if(root->left) cur += '(' + tree2str(root->left) +')';
else if(root->right) cur += "()"; //for no left child but right child case
if(root->right) cur += '(' + tree2str(root->right) +')';
return cur;
}
//Solution B
void solution(TreeNode* root,string &res){
if(!root) return;
res += to_string(root->val);
if(root->left || root->right){
res+= "(";
solution(root->left,res);
res+= ")";
if(root->right){
res+= "(";
solution(root->right,res);
res+= ")";
}
}
}
};