LeetCode 156. Binary Tree Upside Down

原题链接在这里：https://leetcode.com/problems/binary-tree-upside-down/

题目：

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  

题解：

Recursion 方法是自底向上.

Time Complexity: O(n).

Space: O(n). tree height.

AC Java:

/**
 * 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;
    }
}

Iterative 是从上到下.

Time Complexity: O(n). Space: O(1).

AC Java:

/**
 * 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 cur = root;
        TreeNode next = null;
        TreeNode pre = null;
        TreeNode temp = null;
        while(cur != null){
            next = cur.left;
            cur.left = temp;
            temp = cur.right;
            cur.right = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
}

类似Reverse Linked List.