题解 | #牛群的树形结构重建II#

/**
 * 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 preOrder int整型vector 
     * @param inOrder int整型vector 
     * @return TreeNode类
     */
    int find_index(vector<int>& preOrder, vector<int>& inOrder) {
        for (int i = 0; i< inOrder.size(); i++) {
            if (preOrder[0] == inOrder[i]) {
                return i;
            }
        }
        return 0;
    }
    TreeNode* buildTreeII(vector<int>& preOrder, vector<int>& inOrder) {
        // write code here
        if (inOrder.size() == 0) return nullptr;
        TreeNode* root = new TreeNode(preOrder[0]);
        int index = find_index(preOrder, inOrder);
        vector<int> pre = vector<int>(preOrder.begin()+1, preOrder.begin()+index+1);
        vector<int> in = vector<int>(inOrder.begin(),inOrder.begin()+index);
        root->left = buildTreeII(pre, in);
        vector<int> pre1 = vector<int>(preOrder.begin()+index+1, preOrder.end());
        vector<int> in1 = vector<int>(inOrder.begin()+index+1,inOrder.end());
        root->right = buildTreeII(pre1, in1);
        return root;
    }
};