[LeetCode] 114. Flatten Binary Tree to Linked List Java
阿新 • • 發佈:2017-07-24
ini like lin ont == lis example 遞歸 font
題目:
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
題意及分析:給出一棵樹,要求給出這棵樹的先序遍歷組成的鏈表,但是還是用樹表示。首先找到根節點左子節點的最右子節點,然後將根節點的右子樹移到該點的右節點上;再將根節點的左子節點移到根節點的右子節點上,並將根節點左子樹重置為null;然後將根節點向右子節點移動一位,遞歸即可。
代碼:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public void flatten(TreeNode root) { if(root==null) return; if(root.left!=null){ TreeNode cur = root.left; while(cur.right!=null){ cur=cur.right; } cur.right=root.right; root.right=root.left; root.left=null; } flatten(root.right); } }
[LeetCode] 114. Flatten Binary Tree to Linked List Java