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156. Binary Tree Upside Down

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;            }}

 

156. Binary Tree Upside Down