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
一刷
題解：只保留右子樹斤程。我們先用right-left-mid的順序遍歷，把前一個(gè)訪問的用prev存起來橡伞。
private TreeNode prev = null;
public void flatten(TreeNode root) {
if (root == null)
return;
flatten(root.right);
flatten(root.left);
root.right = prev;
root.left = null;
prev = root;
}
二刷：
按照right-left-mid的順序遍歷這棵樹檩赢。那么對于新的root, prev為就會是mid-left-right這種preorder昼蛀， root的右子樹為prev, 左子樹為null
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private TreeNode prev = null;
public void flatten(TreeNode root) {
//preorder
if(root == null) return;
flatten(root.right);
flatten(root.left);
root.left = null;
root.right = prev;
prev = root;
}
}
三刷
post order traverse : right, left, root
root.right = prev;
root.left = null;
prev = root;
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private TreeNode prev = null;
public void flatten(TreeNode root) {
if(root == null) return;
flatten(root.right);
flatten(root.left);
root.right = prev;
root.left = null;
prev = root;
}
}
