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：

屏幕快照 2017-09-06 下午9.28.14.png

Solution1：递归实现

Time Complexity: O(N) Space Complexity: O(N) 递归缓存

Solution2：Iterative实现

Time Complexity: O(N) Space Complexity: O(1)

Solution1 Code：

class Solution1 {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        if(root == null || root.left == null) {
            return root;
        }

        TreeNode newRoot = upsideDownBinaryTree(root.left);
        root.left.left = root.right;  // 1st change
        root.left.right = root;     // 2nd change
        root.left = null;
        root.right = null;
        return newRoot;
    }
}

Solution2 Code：

class Solution2 {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        if(root == null) return null;
        
        TreeNode cur = root;
        TreeNode prev = null;
        TreeNode save_prev_right = null;
        
        while(cur != null) {
            TreeNode next = cur.left;
            
            cur.left = save_prev_right;
            save_prev_right = cur.right;
            cur.right = prev;
            
            prev = cur;
            cur = next;
        }
        return prev;
    }
}