114. Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

    1
   / 
  2   5
 /    
3   4   6

The flattened tree should look like:

1
 
  2
   
    3
     
      4
       
        5
         
          6

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    //要求123456，那么recursion就是654321
    //如何遍历得到654321？right->left->root即可
    private TreeNode prev = null;
    public void flatten(TreeNode root) {
        if(root == null) return;
        flatten(root.right);
        flatten(root.left);
        root.right = prev;
        root.left = null;
        prev = root;
    }
}

有意思

https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/36977/My-short-post-order-traversal-Java-solution-for-share

https://www.cnblogs.com/springfor/p/3864355.html

2020/7/10，我垃圾，看不懂递归，但能看懂stack做法（在大佬的帮助下

 https://www.youtube.com/watch?v=v2ob-ek9TgE

class Solution {
    public void flatten(TreeNode root) {
        if(root == null) return;
        Stack<TreeNode> stack = new Stack();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode cur = stack.pop();
            if(cur.right != null) stack.push(cur.right);
            if(cur.left != null) stack.push(cur.left);
            if(!stack.isEmpty()){
                cur.right = stack.peek();
            }
            cur.left = null;
        }
    }
}

用stack来保证永远优先处理left node，先把rootpush进去，然后在循环里pop栈顶，加入right再加入left，因为用的时候先用left。

如果栈不为空的话，就让cur.right = stack.peek(),这样就能让cur优先指向left node了

然后继续循环直到结束