LeetCode 114 Flatten Binary Tree to Linked List

题目

---

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

思路

---

1 能够想到这其实是中序遍历的变形，就成功一半了。

2 一个节点左边的遍历好，然后处理一下，再遍历右边。

3处理的部分为：这个节点的左边设为null，右边接左边处理好的内容，再接原右边。

4 逻辑清楚了，代码自然也就清楚。

代码

---

public class Solution {    public void flatten(TreeNode root) {        if(root ==null){            return ;        }        TreeNode oldleft = root.left;        TreeNode oldright = root.right;        root.left = null;        flatten(left);        if(left!=null){            root.right = left;            TreeNode cur = left;            while(cur.right!=null){                cur = cur.right;            }            cur.right = right;        }        flatten(right);    }}

15.4.14

可以用递归算法，更加简洁。看代码就能够理解意思

public class Solution {    public void flatten(TreeNode root) {        if(root == null){            return;        }        while(root!=null){            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;            }            root = root.right;        }    }}