[105]从前序与中序遍历序列构造二叉树

//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 
// Related Topics 树 深度优先搜索 数组 
// 👍 893 👎 0

/*
* 105 从前序与中序遍历序列构造二叉树
* 2021-02-27 11:45:39
* @author oxygenbytes
*/ 
#include "leetcode.h" 
//leetcode submit region begin(Prohibit modification and deletion)
/**
 * 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 {
public:
    unordered_map<int, int> mp;
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int n = preorder.size();
        if(!n) return nullptr;

        for(int i = 0;i < inorder.size();i++) mp[inorder[i]] = i;

        return dfs(preorder, inorder, 0, n - 1, 0 ,n - 1);
    }

    TreeNode* dfs(vector<int>& preorder, vector<int>& inorder, int pl, int pr, int il, int ir){
        if (pl > pr) return nullptr;
        int val = preorder[pl];
        int k = mp[val];
        int len = k - il;
        auto root = new TreeNode(val);

        root->left = dfs(preorder, inorder, pl + 1, pl + len, il, k - 1);
        root->right = dfs(preorder, inorder, pl + len + 1, pr, k + 1, ir);

        return root;
    }
};
//leetcode submit region end(Prohibit modification and deletion)