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) {}
 * };
 */
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* subTree(vector<int>& preOrder, vector<int>& vinOrder, int p1, int p2, int i1, int i2)
    {
        if(p1>p2)
            return nullptr;
        
        // 分配新的TreeNode对象
        TreeNode* root = new TreeNode(preOrder[p1]);
        int i=i1, k=0;
	  	// 确定左子树和右子树的片段
        while(vinOrder[i]!=preOrder[p1]) i++, k++;
        root->left = subTree(preOrder, vinOrder, p1+1, p1+k, i1, i-1);
        root->right = subTree(preOrder, vinOrder, p1+k+1, p2, i+1, i2);
        return root;
    }

    TreeNode* reConstructBinaryTree(vector<int>& preOrder, vector<int>& vinOrder) {
        if(preOrder.empty() || vinOrder.empty() || preOrder.size() != vinOrder.size())
            return nullptr;
        return subTree(preOrder, vinOrder, 0, preOrder.size()-1, 0, vinOrder.size()-1);
    }
};