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
Hints:
If you notice carefully in the flattened tree, each node’s right child points to the next node of a pre-order traversal.
题解:可以看出flatten以后得到的树其实就是原来树的先序遍历,并且所得到的树的左子都有null,右子都是原树先序遍历时的下一个节点。
利用递归先序遍历树即可,用lastNode保存上一个节点的信息,并且在递归过程中要注意保存根节点的右子,因为在递归遍历左子树的过程中,会把根节点的右子指针重新赋值,右子树会丢失。
代码如下:
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private TreeNode lastNode = null;
public void flatten(TreeNode root) {
if(root == null)
return; if(lastNode != null){
lastNode.left = null;
lastNode.right = root;
} lastNode = root; TreeNode right = root.right;
flatten(root.left);
flatten(right);
}
}