题解 | #从中序与后序遍历序列构造二叉树#

/**
 * 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 inorder int整型vector 中序遍历序列
     * @param postorder int整型vector 后序遍历序列
     * @return TreeNode类
     */
    void BuildNode(TreeNode *& root,vector<int> inorder, vector<int> postorder,int l1,int r1,int l2,int r2){
        if(r1 < l1 || r2 < l2)
            return ;
        if(root == NULL){
            root = new TreeNode(postorder[r2]);
            root->left = NULL;
            root->right = NULL;
        }
        int pos = find(inorder.begin(),inorder.end(),postorder[r2]) - inorder.begin();
        BuildNode(root->right, inorder, postorder,pos + 1,r1,r2 - (r1 - pos),r2 - 1);
        BuildNode(root->left, inorder, postorder,l1,pos - 1,l2,r2 - (r1 - pos) - 1);
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        // write code here
        TreeNode* root = NULL;
        BuildNode(root, inorder, postorder, 0, inorder.size() - 1, 0,postorder.size() - 1);
        return root;
    }
};