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) {}
 * };
 */
#include <vector>
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* build (vector<int>& inorder,
                     vector<int>& postorder,
                     int ll, int lr,
                     int rl, int rr) {
        if (ll > lr) {
            return nullptr;
        }

        int val = postorder[rr];
        int i = ll;
        while (inorder[i] != val) {
            ++i;
        }

        int size = i - ll;

        TreeNode* root = new TreeNode(val);
        root->left = build(inorder, postorder, ll, i - 1, rl, rl + size - 1);
        root->right = build(inorder,   postorder, i + 1, lr, rl + size, rr - 1);

        return root;
    }

    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        int n = inorder.size();
        return build(inorder, postorder, 0, n - 1, 0, n - 1);
    }
};