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;
}
}