LeetCode 1028: Recover a Tree From Preorder Traversal Solution

Master LeetCode problem 1028 (Recover a Tree From Preorder Traversal), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1028. Recover a Tree From Preorder Traversal

Hard

String Tree Depth-First Search Binary Tree

Problem Explanation

Explanation

To recover a binary tree from the given preorder traversal string, we can use a stack to keep track of the nodes and their depths. We iterate over the characters in the traversal string, extract the node values and depths, and construct the tree accordingly.

The algorithm involves the following steps:

Initialize an empty stack to store nodes and their depths.
Iterate over the traversal string character by character.
Extract the node value and depth from the current substring.
Create the node with the extracted value.
Update the tree structure based on the depth of the current node and the stack.
Push the node and its depth onto the stack for future reference.

At the end of the process, the root node of the recovered tree will be at the top of the stack.

Time complexity: O(n), where n is the length of the traversal string. Space complexity: O(n), where n is the length of the traversal string.

Solution Code

class Solution {
    public TreeNode recoverFromPreorder(String traversal) {
        Stack<TreeNode> stack = new Stack<>();
        int i = 0;
        
        while (i < traversal.length()) {
            int depth = 0;
            while (i < traversal.length() && traversal.charAt(i) == '-') {
                depth++;
                i++;
            }
            
            int value = 0;
            while (i < traversal.length() && Character.isDigit(traversal.charAt(i))) {
                value = value * 10 + (traversal.charAt(i) - '0');
                i++;
            }
            
            TreeNode node = new TreeNode(value);
            
            while (stack.size() > depth) {
                stack.pop();
            }
            
            if (!stack.isEmpty()) {
                TreeNode parent = stack.peek();
                if (parent.left == null) {
                    parent.left = node;
                } else {
                    parent.right = node;
                }
            }
            
            stack.push(node);
        }
        
        while (stack.size() > 1) {
            stack.pop();
        }
        
        return stack.peek();
    }
}

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