给定一个二叉树,原地将它展开为链表。
例如,给定二叉树
1
/ \
2 5
/ \ \
3 4 6
将其展开为:
1
\
2
\
3
\
4
\
5
\
6
思路:采用后序遍历,将左子树放在右子树的位置上,然后再把右子树跟在左子树最右节点的右子树上。
public class Flatten {
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
public void flatten(TreeNode root) {
if (root == null) {
return;
}
flatten(root.left);
flatten(root.right);
if (root.left != null) {
TreeNode left = root.left;
TreeNode right = root.right;
root.right = root.left;
root.left = null;
while (left.right != null) {
left = left.right;
}
left.right = right;
}
}
}