题解 | #重建二叉树#

/**
 * struct TreeNode {
 *  int val;
 *  struct TreeNode *left;
 *  struct TreeNode *right;
 *  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <vector>
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param preOrder int整型vector
     * @param vinOrder int整型vector
     * @return TreeNode类
     */
    TreeNode* reConstructBinaryTreeFun(vector<int>& preOrder, vector<int>& vinOrder,
                                       int preBegin, int preEnd, int vinBegin, int vinEnd) {
        if (preOrder.empty() || vinOrder.empty())
            return nullptr;
        auto root = new TreeNode(preOrder[preBegin]);
        if (preBegin == preEnd)
            return root;
        int index = vinBegin;
        while (index != vinEnd + 1 && vinOrder[index] != preOrder[preBegin]) {
            index++;
        }
        if (index == vinBegin) {
            root->right = reConstructBinaryTreeFun(preOrder, vinOrder, preBegin + 1, preEnd,
                                                   vinBegin + 1, vinEnd);
            return root;
        }
        if (index == vinEnd) {
            root->left = reConstructBinaryTreeFun(preOrder, vinOrder, preBegin + 1, preEnd,
                                                  vinBegin, vinEnd - 1);
            return root;
        }
        root->left = reConstructBinaryTreeFun(preOrder, vinOrder, preBegin + 1,
                                              preBegin + index - vinBegin, vinBegin, index - 1);
        root->right = reConstructBinaryTreeFun(preOrder, vinOrder,
                                               preBegin + index - vinBegin + 1, preEnd, index + 1, vinEnd);
        return root;
    }
    TreeNode* reConstructBinaryTree(vector<int>& preOrder, vector<int>& vinOrder) {
        // write code here
        if (preOrder.empty() || vinOrder.empty())
            return nullptr;
        return reConstructBinaryTreeFun(preOrder, vinOrder, 0, preOrder.size() - 1, 0,
                                        vinOrder.size() - 1);
    }
};