LeetCode-114. Flatten Binary Tree to Linked List
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题目描述
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
解题思路
采用递归的方式求解,对于每一个节点root,保存该节点的右节点到临时变量,对左右节点分别递归调用。先处理左节点,并将处理的结果作为该root节点的右节点(这就是前面为什么保存右节点的原因),然后找到当root节点的右节点的尾部tmp,处理右节点,并将结果作为tmp的右节点,最后返回root节点。
代码
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */class Solution { public void flatten(TreeNode root) { if(root == null) return; root = flatternCore(root); // Queue<TreeNode> queue = new LinkedList<TreeNode>(); // while() } private TreeNode flatternCore(TreeNode root){ TreeNode right = root.right; root.right = null;//必须清空右节点,如果左节点不为空会将右节点覆盖,但是如果左节点为空,则会在计算root的右节点时出问题。 if(root.left != null){ root.right = flatternCore(root.left); root.left = null; } if(right != null){ TreeNode tmp = root; while(tmp.right != null){ tmp=tmp.right; } tmp.right=flatternCore(right); } return root; }}
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