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Invert a binary tree.
Example:
Input:
4 / 2 7 / \ / 1 3 6 9
Output:
4 / 7 2 / \ / 9 6 3 1
Trivia:
This problem was inspired by this original tweet by Max Howell:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.
Problem link
You can find the detailed video tutorial here
Simple question on understanding about recursion. It could be solved using pre-order and post-order traversal style recursion. It can also be solved iteratively using a queue. Please refer to Leetcode official solution. It‘s very similar to "Binary Tree ZigZag Level Order Traversal". Remember a full binary tree max leaf is (N+1)/2 or ceiling of N/2
, that‘s the O(N) space complexity using a queue.
If you are like Max Howell, you can show off but I would still recommend to stay humble if you really want the job. Brillient jerks are rare (IMHO Linus Torvalds is one) but very productive, but not all companies would accept those culture (Netflix does but not others). Be strong and humble.
1 public TreeNode invertTree(TreeNode root) { 2 return this.helper(root); 3 } 4 5 // post order 6 private TreeNode helper(TreeNode root) { 7 if (root == null) { 8 return root; 9 } 10 11 TreeNode l = helper(root.left); 12 TreeNode r = helper(root.right); 13 14 root.left = r; 15 root.right = l; 16 17 return root; 18 } 19 20 // Preorder 21 private TreeNode invertHelper(TreeNode root) { 22 if (root == null) { 23 return null; 24 } 25 26 TreeNode temp = root.left; 27 root.left = root.right; 28 root.right = temp; 29 30 this.invertHelper(root.left); 31 this.invertHelper(root.right); 32 33 return root; 34 }
Time Complexity: O(N), where N is the total number of tree nodes
Space Complexity: O(1) Or O(lgN) if you count the recursion function stack
Leetcode solution 226. Invert Binary Tree
标签:out ril turn original open inpu status ini VID
原文地址:https://www.cnblogs.com/baozitraining/p/12148000.html