题解 | #重建二叉树#

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        if(!pre.size())
            return nullptr;
        int root_ele = pre[0], root_idx_in_vin = 0; // 构建递归的子任务是寻找每个子树的根节点
        for(int i = 0; i < vin.size(); i++){
            if(vin[i] == root_ele){
                root_idx_in_vin = i;
                break;
            }
        }
        vector<int> pre_left, pre_right, vin_left, vin_right;
        for(int i = 0; i < root_idx_in_vin; i++){
            pre_left.push_back(pre[i+1]); // 从非root结点开始
            vin_left.push_back(vin[i]);
        }
        for(int i = root_idx_in_vin+1; i < pre.size(); i++){
            pre_right.push_back(pre[i]);
            vin_right.push_back(vin[i]);
        }
        TreeNode* root = new TreeNode(0); 
        root->val = root_ele;
        root->left = reConstructBinaryTree(pre_left, vin_left);
        root->right = reConstructBinaryTree(pre_right, vin_right);
        return root;
    }
};