LeetCode 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.

Example:

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

    1
   / 
  2   3
 / 
4   5

Output: return the root of the binary tree [4,5,2,#,#,3,1]
翻转以后如下:

   4
  / 
 5   2
    / 
   3   1 

解题思路:
翻转的形式一开始不是很清楚,但是discuss里面的高票答案给了一个很好的解释。看例子,树的左边最深的底层是4,4是新的root。对于每个root node,将链接右孩子的指针去掉,将root node变为当前左孩子的left node,root node成为左孩子的right node。

    1
   /  x
  2 -- 3
 /  x
4 -- 5
^
new root

递归的写法:

    public TreeNode upsideDownBinaryTree(TreeNode root) {
        if (root == null || root.left == null) {
            return root;
        }
        //递归调用得到新的root,并且沿途改变结构。
        TreeNode newRoot = upsideDownBinaryTree(root.left);
        root.left.left = root.right;
        root.left.right = root;
        //千万记得将root node 的左右两边设为null
        root.left = null;
        root.right = null;
        return newRoot;
    }

遍历的解法
遍历的解法需要四个指针,如图所示,每次先update next,然后对swap上一个node的右孩子和这个node的左孩子,所以每次我们需要一个temp来记录上一个node的右边孩子。

   prev -> 1
          /  x
 curr -> 2 -- 3  <-temp
        /  x
next-> 4 -- 5
      ^
    new root

代码如下

    public TreeNode upsideDownBinaryTree(TreeNode root) {
        //iterative
        TreeNode curr = root;
        TreeNode prev = null;
        TreeNode next = null;
        TreeNode temp = null;
        while(curr != null) {
            next = curr.left;
            //swap nodes, we need to keep a temp to track the right node
            curr.left = temp;
            temp = curr.right;
            curr.right = prev;
            
            prev = curr;
            curr = next;
        }
        return prev;

本文固定链接: http://www.js-code.com/node-js/node-js_35943.html