leetcode 156: Binary Tree Upside Down

Binary Tree Upside Down

Total Accepted: 813
Total Submissions: 2515

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

confused what

"{1,#,2,3}"

means?
> read more on how binary tree is serialized on OJ.
[分析]
搞清此树的定义,答案自明.
[注意]
None

[code]/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode UpsideDownBinaryTree(TreeNode root) {
Stack<TreeNode> stack = new Stack<TreeNode>();

if(root==null) return null;
while(root.left != null) {stack.push(root); root=root.left;}
stack.push(root);
while(!stack.empty()) {
TreeNode node = stack.pop();
if(!stack.empty()) {
node.right = stack.peek();
node.left = stack.peek().right;
} else {
node.left = null;
node.right = null;
}
}
return root;
}
}