156. Binary Tree Upside Down

Posted 2020-08-09 Machelsky

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  

 

 思路：不是特别理解怎么翻转，用recursive做的。完全按照题目给的例子来。翻转左子树，然后原来的root.left重新分配指向。

/**
 * 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&&root.right==null)
        {
            return root;
        }
        TreeNode check=upsideDownBinaryTree(root.left);
        root.left.left=root.right;
        root.left.right=root;
        root.left=null;
        root.right=null;
        return check;
        
    }
}