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
     */
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        // write code here
        if(postorder.size()==0)
        return nullptr;
        TreeNode*root=new TreeNode(postorder[postorder.size()-1]);
        for(int i=0;i<inorder.size();i++){
            if(root->val==inorder[i]){
            vector<int> leftin(inorder.begin(), inorder.begin() + i);
            vector<int> rightin(postorder.begin(),postorder.begin()+i);

            // 划分后序遍历的左子树和右子树
            vector<int> leftor(inorder.begin()+i+1,inorder.end());
            vector<int> rightor(postorder.begin() + i, postorder.end() - 1);
                root->left=buildTree(leftin,rightin);
                root->right=buildTree(leftor,rightor);
                return root;
            }

        }
        return root;
    }
};