[LeetCode] 156. Binary Tree Upside Down

Given the root of a binary tree, turn the tree upside down and return the new root.

You can turn a binary tree upside down with the following steps:

The original left child becomes the new root.
The original root becomes the new right child.
The original right child becomes the new left child.

The mentioned steps are done level by level, it is guaranteed that every node in the given tree has either 0 or 2 children.

Example 1:

Input: root = [1,2,3,4,5]
Output: [4,5,2,null,null,3,1]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Constraints:

The number of nodes in the tree will be in the range [0, 10].
1 <= Node.val <= 10
Every node has either 0 or 2 children.

上下翻转二叉树。题意是给一个二叉树，请你翻转一下。我的理解是基本是把这个二叉树顺时针旋转120度的感觉。

如上图，最小的左孩子成了根节点，而原来的根节点成了最小的右孩子。我这里给出递归和迭代两种做法。首先递归的思路是不断往左子树走，找到新的根节点，然后将新的根节点跟他的左右孩子分别连好，然后再断开原来的根节点及其左右孩子的联系。

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public TreeNode upsideDownBinaryTree(TreeNode root) {
18         // corner case
19         if (root == null || root.left == null && root.right == null) {
20             return root;
21         }
22         
23         // normal case
24         TreeNode newRoot = upsideDownBinaryTree(root.left);
25         root.left.left = root.right;
26         root.left.right = root;
27         root.left = null;
28         root.right = null;
29         return newRoot;
30     }
31 }

迭代的做法需要用到stack，有点类似中序遍历的感觉。一开始也是需要不断往左子树走，边走边把遇到的节点入栈。当走到最左子树的时候会停下，此时pop出来的栈顶元素就是新的根节点newRoot。再pop一个元素出来，这个元素是新的根节点的右孩子，因为他也是原来的某个父级节点，所以我们暂时把他命名为oriParent。同时oriParent的右孩子（如果有的话），是newRoot的左孩子。最后记得把oriParent原来的左右链接断开。

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public TreeNode upsideDownBinaryTree(TreeNode root) {
18         // corner case
19         if (root == null) {
20             return null;
21         }
22         // normal case
23         Stack<TreeNode> stack = new Stack<>();
24         while (root != null) {
25             stack.push(root);
26             root = root.left;
27         }
28         TreeNode newRoot = stack.pop();
29         TreeNode cur = newRoot;
30         while (!stack.isEmpty()) {
31             TreeNode oriParent = stack.pop();
32             cur.right = oriParent;
33             if (oriParent.right != null) {
34                 cur.left = oriParent.right;
35             }
36             cur = oriParent;
37             oriParent.left = null;
38             oriParent.right = null;
39         }
40         return newRoot;
41     }
42 }

LeetCode 题目总结