[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:
1
/ \
2 5
/ \ \
3 4 6
1
/ \
2 5
\ \
3 6
\
4
1
\
2
\
3
\
4
\
5
\
6
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:
1
/ \
2 5
/ \ \
3 4 6
1
\
2
/ \
3 4
\
5
\
6
1
\
2
\
3
\
4
\
5
\
6
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);
}
}
};
此题还可以延伸到用中序,后序,层序的遍历顺序来展开原二叉树,分别又有其对应的递归和非递归的方法,有兴趣的童鞋可以自行实现。
Github 同步地址:
https://github.com/grandyang/leetcode/issues/114
类似题目:
Flatten a Multilevel Doubly Linked List
参考资料:
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/37182/my-recursive-solution-is-easy-and-clean
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/36977/my-short-post-order-traversal-java-solution-for-share
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