[力扣题解] 106. 从中序与后序遍历序列构造二叉树

题目：106. 从中序与后序遍历序列构造二叉树

思路

递归
坚持用左闭右开划分区间；

代码

/**
 * 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:
    // [in_start, end)
    TreeNode* travel(vector<int>& inorder, int in_start, int in_end,
                vector<int>& postorder, int post_start, int post_end)
    {
        if(post_start == post_end)
        {
            return NULL;
        }

        int root_value;
        int mid;
        int in_left_start, in_left_end, in_right_start, in_right_end;
        int post_left_start, post_left_end, post_right_start, post_right_end;
        root_value = postorder[post_end - 1];
        TreeNode* root = new TreeNode(root_value);

        if(postorder.size() == 1)
        {
            return root;
        }

        for(mid = in_start; mid < in_end; mid++)
        {
            if(inorder[mid] == root_value)
            {
                break;
            }
        }

        // 切割中序遍历
        // [in_start, mid)
        in_left_start = in_start;
        in_left_end = mid;
        // 假设 in_start = 0, -> [0, mid)
        // 两边同时加上 post_start
        // [post_start, post_start + mid)
        // [mid+1, in_end)
        in_right_start = mid+1;
        in_right_end = in_end;

        // 切割后序遍历
        // [in_start, post_start + (mid - in_start))
        post_left_start = post_start;
        post_left_end = post_start + (mid - in_start);
        // [post_start + (mid - in_start), post_end - 1)
        post_right_start = post_start + (mid - in_start);
        post_right_end = post_end - 1;

        root->left = travel(inorder, in_left_start, in_left_end, postorder, post_left_start, post_left_end);
        root->right = travel(inorder, in_right_start, in_right_end, postorder, post_right_start, post_right_end);
        return root;
    }

    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if(inorder.size() == 0 && postorder.size() == 0)
        {
            return NULL;
        }
        // [0, inorder,size())
        // [0, postorder.size())
        return travel(inorder, 0, inorder.size(), postorder, 0, postorder.size());
    }
};