《Cracking the Coding Interview》——第17章：普通题——题目13

2014-04-29 00:15

题目：将二叉搜索树展开成一个双向链表，要求这个链表仍是有序的，而且不能另外分配对象，就地完成。

解法：Leetcode上也有，递归解法。

代码：

// 17.13 Flatten a binary search tree into a doubly linked list by inorder traversal order.
// Use postorder traversal to do the flattening job.
#include <cstdio>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int _val = 0): val(_val), left(nullptr), right(nullptr) {};
};

void flatten(TreeNode *&root, TreeNode *&left_most, TreeNode *&right_most)
{
    if (root == nullptr) {
        left_most = right_most = nullptr;
        return;
    }
    
    TreeNode *ll, *lr, *rl, *rr;
    if (root->left != nullptr) {
        flatten(root->left, ll, lr);
        root->left = lr;
        lr->right = root;
    } else {
        ll = lr = root;
    }
    
    if (root->right != nullptr) {
        flatten(root->right, rl, rr);
        root->right = rl;
        rl->left = root;
    } else {
        rl = rr = root;
    }
    
    left_most = ll;
    right_most = rr;
}

void constructBinaryTree(TreeNode *&root)
{
    int val;
    
    if (scanf("%d", &val) != 1) {
        root = nullptr;
    } else if (val == 0) {
        root = nullptr;
    } else {
        root = new TreeNode(val);
        constructBinaryTree(root->left);
        constructBinaryTree(root->right);
    }
}

void deleteList(TreeNode *&head)
{
    TreeNode *ptr;
    
    while (head != nullptr) {
        ptr = head;
        head = head->right;
        delete ptr;
    }
}

int main()
{
    TreeNode *root;
    TreeNode *left_most, *right_most;
    TreeNode *head;
    TreeNode *ptr;
    
    while (true) {
        constructBinaryTree(root);
        if (root == nullptr) {
            break;
        }
        flatten(root, left_most, right_most);
        head = left_most;
        for (ptr = head; ptr != nullptr; ptr = ptr->right) {
            printf("%d ", ptr->val);
        }
        putchar('\n');
        deleteList(head);
    }
    
    return 0;
}