Description

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Pick One

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.

Example:

Input: [1,2,3,4,5]

image.png

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

image.png

Clarification:

Confused what [4,5,2,#,#,3,1] means? Read more below on how binary tree is serialized on OJ.

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

image.png

The above binary tree is serialized as [1,2,3,#,#,4,#,#,5].

Solution

Recursion, time O(n), space O(n)

研究了好半天才明白这道题是什么意思....

首先这棵树是左倾斜的，呈一把梳子的形状，然后题目需要把树顺时针翻转，变成这样：

image.png

对左子树进行递归即可，因为右子树要么为空要么是叶子节点。

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

Iteration, time O(n), space O(1)

image.png

图中的temp我这里命名为right，更好理解。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        TreeNode prev = null;
        TreeNode curr = root;
        TreeNode right = null;
        
        while (curr != null) {
            TreeNode next = curr.left;
            curr.left = right;
            right = curr.right;
            curr.right = prev;
            prev = curr;
            curr = next;
        }
        
        return prev;
    }
}