LeetCode——Flatten Binary Tree to Linked List
2015-08-07 12:42
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Given a binary tree, flatten it to a linked list in-place.
For example,
Given
The flattened tree should look like:
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。
reference : http://blog.csdn.net/perfect8886/article/details/20000083
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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