LeetCode - 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：

起始对于每一个节点，相应的操作为：
p.left = parent.right;
p.right = parent;

 

public class Solution {
    public TreeNode upsideDownBinaryTree(TreeNode root) {
        TreeNode p = root, parent = null, parentRight = null;
        while (p!=null) {
            TreeNode left = p.left;
            p.left = parentRight;
            parentRight = p.right;
            p.right ＝ parent;
            parent = p;
            p = left;
        }
        return parent;
    }
}

 

Bottom up approach:

Although the code for the top-down approach seems concise, it is actually subtle and there are a lot of hidden traps if you are not careful. The other approach is thinking recursively in a bottom-up fashion. If we reassign the bottom-level nodes before the upper ones, we won’t have to make copies and worry about overwriting something. We know the new root will be the left-most leaf node, so we begin the reassignment here.

public TreeNode upsideDownBinaryTree(TreeNode root) {
   return dfsBottomUp(root, null);
}
private TreeNode dfsBottomUp(TreeNode p, TreeNode parent) {
   if (p == null) return parent;
   TreeNode root = dfsBottomUp(p.left, p);
   p.left = (parent == null) ? parent : parent.right;
   p.right = parent;
   return root;
}