leetcode(리트코드)105-Construct Binary Tree from Preorder and Inorder Traversal

leetcode 105 - Construct Binary Tree from Preorder and Inorder Traversal 문제입니다.

1. 문제

https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

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2. Input , Output

Constraints:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder and inorder consist of unique values. Each value of inorder also appears in preorder.
• preorder is guaranteed to be the preorder traversal of the tree.
• inorder is guaranteed to be the inorder traversal of the tree.

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3. 분류 및 난이도

Medium 난이도 문제입니다.
leetcode Top 100 Liked 문제입니다.

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4. 문제 해석

• Preorder와 Inorder를 가지고 맞게 트리를 구성해야합니다.
• 재귀라는 것은 알았으나 너무 어려웠습니다. Discuss를 봤습니다.

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5. code

c++

class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int rootIdx = 0;
        return build(preorder, inorder, rootIdx, 0, inorder.size()-1);
    }
    
    TreeNode* build(vector<int>& preorder, vector<int>& inorder, int& rootIdx, int left, int right) {
        if (left > right) return NULL;
        int pivot = left;  // find the root from inorder
        while(inorder[pivot] != preorder[rootIdx]) pivot++;
        
        rootIdx++;
        TreeNode* newNode = new TreeNode(inorder[pivot]);
        newNode->left = build(preorder, inorder, rootIdx, left, pivot-1);
        newNode->right = build(preorder, inorder, rootIdx, pivot+1, right);
        return newNode;
    }
};

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python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def buildTree(self, preorder, inorder):
        if inorder:
            ind = inorder.index(preorder.pop(0))
            root = TreeNode(inorder[ind])
            root.left = self.buildTree(preorder, inorder[0:ind])
            root.right = self.buildTree(preorder, inorder[ind+1:])
            return root
        

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6. 결과 및 후기, 개선점

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