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) {}
 * };
 {1,2,4,7,3,5,6,8}
 {4,7,2,1,5,3,8,6}
 */
#include <vector>
class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        int m = pre.size();
        int n = vin.size();
        if(m == 0 || n == 0){
            return NULL;
        }

        TreeNode* root = new TreeNode(pre[0]);
        for(int i=0;i<pre.size();i++){
            if(pre[0] == vin[i]){
                //左子树前序遍历
                vector<int >leftpre(pre.begin()+1, pre.begin()+i+1);//end指向最后一个元素后一位置
                //左子树中序遍历
                vector<int >leftvin(vin.begin(), vin.begin()+i);
                root->left = reConstructBinaryTree(leftpre, leftvin);
                //右子树前序遍历
                vector<int >rightpre(pre.begin()+i+1, pre.end());
                //右子树中序遍历
                vector<int >rightvin(vin.begin()+i+1, vin.end());
                root->right = reConstructBinaryTree(rightpre, rightvin);
                break;
            }
           
        }
        return root;
    }
};