给定一个二叉树,使用原地算法将它 “压扁” 成链表。
示例:
给出:
         1
        / \
       2   5
      / \   \
     3   4   6
压扁后变成如下:
   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6
提示:
如果您细心观察该扁平树,则会发现每个节点的右侧子节点是以原二叉树前序遍历的次序指向下一个节点的。

详见:https://leetcode.com/problems/flatten-binary-tree-to-linked-list/description/

Java实现:

递归实现:

/**
* 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;
}
if(root.left!=null){
flatten(root.left);
}
if(root.right!=null){
flatten(root.right);
}
TreeNode tmp=root.right;
root.right=root.left;
root.left=null;
while(root.right!=null){
root=root.right;
}
root.right=tmp;
}
}

非递归实现:

/**
* 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;
}
TreeNode cur=root;
while(cur!=null){
if(cur.left!=null){
TreeNode p=cur.left;
while(p.right!=null){
p=p.right;
}
p.right=cur.right;
cur.right=cur.left;
cur.left=null;
}
cur=cur.right;
}
}
}

python实现:

参考:https://www.cnblogs.com/grandyang/p/4293853.html

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