C++实现LeetCode(114.将二叉树展开成链表)

[LeetCode] 114. Flatten Binary Tree to Linked List 将二叉树展开成链表

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

1
/ \
2   5
/ \   \
3   4   6

The flattened tree should look like:

1
\
2
\
3
\
4
\
5
\
6

click to show hints.

Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order trave

这道题要求把二叉树展开成链表，根据展开后形成的链表的顺序分析出是使用先序遍历，那么只要是数的遍历就有递归和非递归的两种方法来求解，这里我们也用两种方法来求解。首先来看递归版本的，思路是先利用 DFS 的思路找到最左子节点，然后回到其父节点，把其父节点和右子节点断开，将原左子结点连上父节点的右子节点上，然后再把原右子节点连到新右子节点的右子节点上，然后再回到上一父节点做相同操作。代码如下：

解法一：

									class Solution {

									public:

									    void flatten(TreeNode *root) {

									        if (!root) return;

									        if (root->left) flatten(root->left);

									        if (root->right) flatten(root->right);

									        TreeNode *tmp = root->right;

									        root->right = root->left;

									        root->left = NULL;

									        while (root->right) root = root->right;

									        root->right = tmp;

									    }

									};

例如，对于下面的二叉树，上述算法的变换的过程如下：

下面再来看非迭代版本的实现，这个方法是从根节点开始出发，先检测其左子结点是否存在，如存在则将根节点和其右子节点断开，将左子结点及其后面所有结构一起连到原右子节点的位置，把原右子节点连到元左子结点最后面的右子节点之后。代码如下：

解法二：

									class Solution {

									public:

									    void flatten(TreeNode *root) {

									        TreeNode *cur = root;

									        while (cur) {

									            if (cur->left) {

									                TreeNode *p = cur->left;

									                while (p->right) p = p->right;

									                p->right = cur->right;

									                cur->right = cur->left;

									                cur->left = NULL;

									            }

									            cur = cur->right;

									        }

									    }

									};

例如，对于下面的二叉树，上述算法的变换的过程如下：

前序迭代解法如下：

解法三：

									class Solution {

									public:

									    void flatten(TreeNode* root) {

									        if (!root) return;

									        stack<TreeNode*> s;

									        s.push(root);

									        while (!s.empty()) {

									            TreeNode *t = s.top(); s.pop();

									            if (t->left) {

									                TreeNode *r = t->left;

									                while (r->right) r = r->right;

									                r->right = t->right;

									                t->right = t->left;

									                t->left = NULL;

									            }

									            if (t->right) s.push(t->right);

									        }

									    }

									};