105.从前序与中序遍历序列构造二叉树.cpp

#include <vector>
using namespace std;
/*
 * @lc app=leetcode.cn id=105 lang=cpp
 *
 * [105] 从前序与中序遍历序列构造二叉树
 *
 * https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/
 *
 * algorithms
 * Medium (62.87%)
 * Likes:    287
 * Dislikes: 0
 * Total Accepted:    33.3K
 * Total Submissions: 53K
 * Testcase Example:  '[3,9,20,15,7]\n[9,3,15,20,7]'
 *
 * 根据一棵树的前序遍历与中序遍历构造二叉树。
 * 
 * 注意:
 * 你可以假设树中没有重复的元素。
 * 
 * 例如，给出
 * 
 * 前序遍历 preorder = [3,9,20,15,7]
 * 中序遍历 inorder = [9,3,15,20,7]
 * 
 * 返回如下的二叉树：
 * 
 * ⁠   3
 * ⁠  / \
 * ⁠ 9  20
 * ⁠   /  \
 * ⁠  15   7
 * 
 */
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};
// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        TreeNode* root = create(0, inorder.size()-1, 0, preorder.size()-1, preorder, inorder);
        return root;
    }
    TreeNode* create(int inL, int inR, int preL, int preR, vector<int>& preorder, vector<int>& inorder) {
        if (preL > preR) return nullptr;
        
        int rootVal = preorder[preL];
        TreeNode *root = new TreeNode(rootVal);

        int k = 0;
        for (int i=inL; i<=inR; i++) {
            if (rootVal == inorder[i]) {
                k = i;
                break;
            }
        }
        int numLeft = k- inL;  //
        
        root->left = create(inL, k-1, preL+1, preL+numLeft, preorder, inorder);
        root->right = create(k+1, inR, preL+numLeft+1, preR, preorder, inorder);

        return root;
    }
};
// @lc code=end