105. Construct Binary Tree from Preorder and Inorder Traversal

Hint

/**
 * 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 {
private:
    // rootIndex is used to represent the current root node index in preorder
    
    // indexMap is used to store the index of each node in inorder

public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder)
    {
        // Store each value and its index in inorder into indexMap
        
        // Call the recursive function to build the binary tree and return the root node
    }
    
    // build function is used to recursively build the subtree
    TreeNode* build(vector<int>& preorder, int left, int right)
    {
        // If left > right, it indicates that the current interval is invalid, return a null node
        
        // Get the value of the root node according to preorder[rootIndex] and move rootIndex forward
        
        // Create the root node
        
        // Find the index position of the root node in inorder to divide the left and right subtrees
        
        // Recursively build the left subtree, the interval of the left subtree is [left, inorderIndex - 1]
        
        // Recursively build the right subtree, the interval of the right subtree is [inorderIndex + 1, right]
        
        // Return the root node to complete the construction of the current subtree
    }
};

Solution - Recursion

/**
 * 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 {
private:
    int rootIndex = 0;
    unordered_map<int, int> indexMap;    
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder)
    {
        int n = preorder.size();

        for(int i = 0; i < n; ++i)
        {
            indexMap[inorder[i]] = i;
        }
        return build(preorder, 0, n - 1);
    }
    
    TreeNode* build(vector<int>& preorder, int left, int right)
    {
        if(left > right) return nullptr;
        
        int rootValue = preorder[rootIndex++];
        
        TreeNode* root = new TreeNode(rootValue);

        root->left = build(preorder, left, indexMap[rootValue] - 1);
        
        root->right = build(preorder, indexMap[rootValue] + 1, right);
        return root;
    }
};

• 時間複雜度：
  $O(N)$
• 空間複雜度：
  $O(N)$