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 <iterator>
#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* reConstructBinaryTree(vector<int>& preOrder, vector<int>& vinOrder) {
        // write code here
        if( preOrder.size() != vinOrder.size() || preOrder.size() == 0 ) return nullptr;
        if( preOrder.size() == 1 ) return new TreeNode(preOrder[0]) ;
        for ( int i = 0; i < vinOrder.size(); i++ ){
            if ( vinOrder[i] == preOrder[0] ){
                auto tree_node = new TreeNode(preOrder[0]);
                vector<int> v1(preOrder.begin() + 1 ,preOrder.begin() + i + 1 );
                vector<int> v2(vinOrder.begin() ,vinOrder.begin() + i  );
                tree_node->left = reConstructBinaryTree( v1,v2 );
                vector<int> v3(preOrder.begin() + i + 1 ,preOrder.end() );
                vector<int> v4(vinOrder.begin() + i + 1 ,vinOrder.end() );
                tree_node->right = reConstructBinaryTree( v3,v4 );
                return tree_node;
            } 
        }
        return nullptr;
    }
};