【leetcode刷题笔记】Flatten Binary Tree to Linked List

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

Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

---

 

题解：可以看出flatten以后得到的树其实就是原来树的先序遍历，并且所得到的树的左子都有null，右子都是原树先序遍历时的下一个节点。

利用递归先序遍历树即可，用lastNode保存上一个节点的信息，并且在递归过程中要注意保存根节点的右子，因为在递归遍历左子树的过程中，会把根节点的右子指针重新赋值，右子树会丢失。

代码如下：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private TreeNode lastNode = null;
    public void flatten(TreeNode root) {
        if(root == null)
            return;
        
        if(lastNode != null){
            lastNode.left = null;
            lastNode.right = root;
        }
        
        lastNode = root;
        
        TreeNode right = root.right;
        flatten(root.left);
        flatten(right);
    }
}