链接

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* constructFromPrePost(vector<int>& preorder, vector<int>& postorder) {
        if (preorder.size() == 0) {
            return nullptr;
        }
        int val = preorder[0];
        TreeNode *node = new TreeNode(val);
        if (preorder.size() == 1) {
            return node;
        }
        vector<int> prev1;
        vector<int> postv1;
        vector<int> prev2;
        vector<int> postv2;
        int tmp = preorder[1];
        for (int i = 0; i < postorder.size()-1; i++) {
            if (postorder[i] != tmp) {
                prev1.push_back(preorder[i+1]);
                postv1.push_back(postorder[i]);
            } else {
                prev1.push_back(preorder[i+1]);
                postv1.push_back(postorder[i]);
                for (int j = i+1; j < postorder.size()-1; j++) {
                    prev2.push_back(preorder[j+1]);
                    postv2.push_back(postorder[j]); 
                }
                break;
            }
        }
        node->left = constructFromPrePost(prev1, postv1);
        node->right = constructFromPrePost(prev2, postv2);
        return node;
    }
};

思路

类似，递归处理，使用最直观的递归方式，时空效率并不高；

参考满二叉树的例子