My code:
/**
* 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 newRoot = upsideDownBinaryTree(root.left);
root.left.left = root.right;
root.left.right = root;
root.left = null;
root.right = null;
return newRoot;
}
}
iteration:
My code:
/**
* 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) {
return root;
}
TreeNode curr = root;
TreeNode pre = null;
TreeNode temp = null;
TreeNode next = null;
while (curr != null) {
next = curr.left;
curr.left = temp;
temp = curr.right;
curr.right = pre;
pre = curr;
curr = next;
}
return pre;
}
}
这道题目都看不懂。。
直到看了代码，自己测试了下，才逐渐懂到底要我们干什么。。
然后所有的右节点，要么是有兄弟结点的叶子结点，要么是null
所以这棵树是 Invalid 的。
[1,2,3,4,5,6,7]
所以只需要考虑一直向左就行。
recursion is from bottom to top
iteration is from top to bottom
Anyway, Good luck, Richardo! -- 09/07/2016
