LeetCode 从前序与中序遍历序列构造二叉树(二叉树)

/**
 * 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* build(vector<int>& preorder, vector<int>& inorder) {
        TreeNode* tree = new TreeNode(preorder[0]);
        if (inorder.size() <= 1) {
            return tree;
        }
        int i, m;
        for (i = 0; i < inorder.size(); i++) {
            if (inorder[i] == preorder[0]) {
                break;
            }
        }
        m = i;
        vector<int> arr0, arr1, brr0, brr1;
        if (m == 0) {
            tree->left = NULL;
        }
        else {
            for (int j = 0, i = 1; j < m; j++, i++) {
                arr0.push_back(preorder[i]);
                brr0.push_back(inorder[j]);
            }
            tree->left = build(arr0, brr0);
        }
        if (m >= inorder.size() - 1) {
            tree->right = NULL;
        }
        else {
            for (int j = m + 1; j < inorder.size(); j++) {
                arr1.push_back(preorder[j]);
                brr1.push_back(inorder[j]);
            }
            tree->right = build(arr1, brr1);
        }
        return tree;
    }

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if (inorder.size() == 0 || preorder.size() == 0) {
            return NULL;
        }
        return build(preorder, inorder);
    }
};

/**
 * 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:
    vector<int> arr, brr;

public:
    TreeNode *build(int a_0, int a_1, int b_0, int b_1)
    {
        TreeNode *tree = new TreeNode(arr[a_0]);
        int i;
        int lefta_0, lefta_1, leftb_0, leftb_1, righta_0, righta_1, rightb_0, rightb_1;
        for (i = b_0; brr[i] != arr[a_0]; i++)
        {
            
        }
        lefta_0 = a_0 + 1;
        lefta_1 = a_0 + i - b_0;
        leftb_0 = b_0;
        leftb_1 = i - 1;
        righta_0 = lefta_1 + 1;
        righta_1 = a_1;
        rightb_0 = i + 1;
        rightb_1 = b_1;
        if (i == b_0)
            tree->left = NULL;
        else
            tree->left = build(lefta_0, lefta_1, leftb_0, leftb_1);
        if (i == b_1)
            tree->right = NULL;
        else
            tree->right = build(righta_0, righta_1, rightb_0, rightb_1);
        return tree;
    }

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if (inorder.size() == 0 || preorder.size() == 0) {
            return NULL;
        }
        arr = preorder;
        brr = inorder;
        return build(0, arr.size() - 1, 0, brr.size() - 1);
    }
};