题解 | #牛群的树形结构重建#

/**
 * struct TreeNode {
 *  int val;
 *  struct TreeNode *left;
 *  struct TreeNode *right;
 *  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
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 inOrder int整型vector
     * @param postOrder int整型vector
     * @return TreeNode类
     */
    TreeNode* buildTree(vector<int>& inOrder, vector<int>& postOrder) {
        int postLen = postOrder.size();
        int inLen = inOrder.size();

        // 处理边界情况
        if (postLen == 0 || inLen == 0 || postLen != inLen) {
            return nullptr;
        }

        // 后序遍历数组的最后一个元素为根节点
        TreeNode* root = new TreeNode(postOrder[postLen - 1]);

        // 在中序遍历数组中找到根节点的位置
        int rootIndex = 0;
        for (int i = 0; i < inLen; i++) {
            if (inOrder[i] == root->val) {
                rootIndex = i;
                break;
            }
        }

        // 构建左子树和右子树的中序遍历数组
        vector<int> leftInorder(inOrder.begin(), inOrder.begin() + rootIndex);
        vector<int> rightInorder(inOrder.begin() + rootIndex + 1, inOrder.end());

        // 构建左子树和右子树的后序遍历数组
        vector<int> leftPostorder(postOrder.begin(), postOrder.begin() + rootIndex);
        vector<int> rightPostorder(postOrder.begin() + rootIndex, postOrder.end() - 1);

        // 递归构造左子树和右子树
        root->left = buildTree(leftInorder, leftPostorder);
        root->right = buildTree(rightInorder, rightPostorder);

        return root;
        // write code here
    }
};