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.
原题链接：https://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/
题目：给定二叉树，按前序位置展平成一个链表。
思路：递归处理。把右子树放到左子树之后，并清空左子树。
public void flatten(TreeNode root) {
if(root == null)
return;
flatten(root.left);
flatten(root.right);
TreeNode tmp = root;
if(tmp.left == null)
return;
else
tmp = tmp.left;
while(tmp.right != null)
tmp = tmp.right;
tmp.right = root.right;
root.right = root.left;
root.left = null;
}
// Definition for binary tree
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
相同的做法。可是非递归。
public void flatten(TreeNode root){
while(root != null){
if(root.left != null){
TreeNode tmp = root.left;
while(tmp.right != null)
tmp = tmp.right;
tmp.right = root.right;
root.right = root.left;
root.left = null;
}
root = root.right;
}
}
reference : http://blog.csdn.net/perfect8886/article/details/20000083
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