Formatted question description: https://leetcode.ca/all/606.html

606. Construct String from Binary Tree (Easy)

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. DFS

Plain DFS. Notice when you should add the left/right child part.

// OJ: https://leetcode.com/problems/construct-string-from-binary-tree

// Time: O(N)
// Space: O(N)
class Solution {
public:
  string tree2str(TreeNode* t) {
    if (!t) return "";
    string left = tree2str(t->left), right = tree2str(t->right);
    string ans = to_string(t->val);
    if (left.size() || right.size())  ans += "(" + left + ")";
    if (right.size()) ans += "(" + right + ")";
    return ans;
  }
};

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public String tree2str(TreeNode t) {
            if (t == null)
                return "";
            if (t.left == null && t.right == null)
                return String.valueOf(t.val);
            StringBuffer sb = new StringBuffer();
            Stack<TreeNode> stack = new Stack<TreeNode>();
            Set<TreeNode> visited = new HashSet<TreeNode>();
            stack.push(t);
            while (!stack.isEmpty()) {
                TreeNode node = stack.peek();
                if (visited.contains(node)) {
                    stack.pop();
                    sb.append(")");
                } else {
                    visited.add(node);
                    sb.append("(");
                    sb.append(node.val);
                    TreeNode left = node.left, right = node.right;
                    if (left == null && right != null)
                        sb.append("()");
                    if (right != null)
                        stack.push(right);
                    if (left != null)
                        stack.push(left);
                }
            }
            String treeStr = sb.toString().substring(1, sb.length() - 1);
            return treeStr;
        }
    }
    
  • // OJ: https://leetcode.com/problems/construct-string-from-binary-tree
    // Time: O(N)
    // Space: O(N)
    class Solution {
    public:
      string tree2str(TreeNode* t) {
        if (!t) return "";
        string left = tree2str(t->left), right = tree2str(t->right);
        string ans = to_string(t->val);
        if (left.size() || right.size())  ans += "(" + left + ")";
        if (right.size()) ans += "(" + right + ")";
        return ans;
      }
    };
    
  • class Solution(object):
      def tree2str(self, t):
        """
        :type t: TreeNode
        :rtype: str
        """
        if not t:
          return ""
        res = ""
        left = self.tree2str(t.left)
        right = self.tree2str(t.right)
        if left or right:
          res += "(%s)" % left
        if right:
          res += "(%s)" % right
        return str(t.val) + res
    
    

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