Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

For example:
Given a binary tree {1,2,3,4,5},

    1
   / \
  2   3
 / \
4   5

return the root of the binary tree [4,5,2,#,#,3,1].

   4
  / \
 5   2
    / \
   3   1  

一刷
题解：

方法1: recursive

recursive

newRoot存在的意义是keep住最left的leaf

public TreeNode upsideDownBinaryTree(TreeNode root) {
    if(root == null || root.left == null) {
        return root;
    }
    
    TreeNode newRoot = upsideDownBinaryTree(root.left);
    root.left.left = root.right;   // node 2 left children
    root.left.right = root;         // node 2 right children
    root.left = null;
    root.right = null;
    return newRoot;
}

方法2： iterative
注意，temp在上一轮中就要保存好，避免被overwritten.

iterative

public TreeNode upsideDownBinaryTree(TreeNode root) {
    TreeNode curr = root;
    TreeNode next = null;
    TreeNode temp = null;
    TreeNode prev = null;
    
    while(curr != null) {
        next = curr.left;
        
        // swapping nodes now, need temp to keep the previous right child
        curr.left = temp;
        temp = curr.right;
        curr.right = prev;
        
        prev = curr;
        curr = next;
    }
    return prev;
}  

二刷
遍历顺序为：left-root

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        if(root == null || root.left == null) return root;
        
        TreeNode newRoot = upsideDownBinaryTree(root.left);
        root.left.left = root.right;
        root.left.right = root;
        root.left = null;
        root.right = null;
        return newRoot;
    }
}