生成二叉树的题目还是有点思考量的，所以在这里存个档

1. 前序遍历+中序遍历。生成二叉树

leetcode版本1

/**
 * 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> pos;
    vector<int> preorder, inorder;
    TreeNode* buildTree(vector<int>& preorder_, vector<int>& inorder_) {
        preorder = preorder_;
        inorder = inorder_;
        int n = inorder_.size();
        for (int i = 0; i < n; i ++ ) pos[inorder_[i]] = i;

        return build(0, n - 1, 0, n - 1);
    }

    TreeNode* build(int il, int ir, int pl, int pr) {
        TreeNode* root = new TreeNode(preorder[pl]);

        int k = pos[root->val];

        // x - (pl + 1) = k - 1 - il;
        if (il < k) root->left = build(il, k - 1, pl + 1, pl + 1 + k - 1 - il);
        if (k < ir) root->right = build(k + 1, ir, pl + 1 + k - 1 - il + 1, pr);
        return root;
    }
};

leetcode 版本2, 这种方法好像更快

/**
 * 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> pos;
    vector<int> preorder, inorder;
    TreeNode* buildTree(vector<int>& preorder_, vector<int>& inorder_) {
        preorder = preorder_;
        inorder = inorder_;
        int n = inorder_.size();
        for (int i = 0; i < n; i ++ ) pos[inorder_[i]] = i;

        return build(0, n - 1, 0, n - 1);
    }

    TreeNode* build(int il, int ir, int pl, int pr) {
        if (pl > pr) return nullptr;

        TreeNode* root = new TreeNode(preorder[pl]);

        int k = pos[root->val];

        // x - (pl + 1) = k - 1 - il;
        root->left = build(il, k - 1, pl + 1, pl + 1 + k - 1 - il);
        root->right = build(k + 1, ir, pl + 1 + k - 1 - il + 1, pr);
        return root;
    }
};

acwing 版本

2. 后序遍历+中序遍历。生成二叉树

leetcode版本

/**
 * 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> pos;
    vector<int> inorder, postorder;

    TreeNode* buildTree(vector<int>& inorder_, vector<int>& postorder_) {
        inorder = inorder_;
        postorder = postorder_;
        for (int i = 0; i < inorder.size(); i ++ ) pos[inorder[i]] = i;

        return build(0, inorder.size() - 1, 0, inorder.size() - 1);
    }

    TreeNode* build(int il, int ir, int pl, int pr) {
        if (pl > pr) return nullptr;

        TreeNode* root = new TreeNode(postorder[pr]);
        int k = pos[root->val];
        // x - pl = k - 1 - il
        root->left = build(il, k - 1, pl, pl + k - 1 - il);
        root->right = build(k + 1, ir, pl + k - 1 - il + 1, pr - 1);
        return root;
    }
};

acwing 版本