Leetcode:Flatten Binary Tree to Linked List

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 traversal.

分析：这道题是pre-order traversal的变种，我们可以很容易的写出recursion的解法，先flatten左子树再flatten右子树，然后将左子树最右节点的right child设为右子树的根。但空间复杂度为O(logn)。代码如下：

class Solution {
public:
void flatten(TreeNode *root) {
if(root == NULL) return;

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

if(root->left == NULL) return;

TreeNode *p = root->left;
for(; p->right; p = p->right);//find right most node in left child tree

p->right = root->right;
root->right = root->left;
root->left = NULL;
}
};

如果要用in place的方法，我们可以联想到Morris遍历。代码如下：

class Solution {
public:
void flatten(TreeNode *root) {
TreeNode *cur = root;

while(cur){
if(cur->left != NULL){
TreeNode *prev = cur->left;
for(; prev->right && prev->right != cur; prev = prev->right);
if(prev->right == NULL){
prev->right = cur;
cur = cur->left;
}else{
prev->right = cur->right;
cur->right = cur->left;
cur->left = NULL;
cur = prev->right;
}
}else{
cur = cur->right;
}
}
}
};