[LintCode] Binary Tree Flipping

时间：2017-09-07 11:59:59 阅读：150 评论：0 收藏：0 [点我收藏+]

标签：des toggle linked time example ges hang win dep

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.

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

Think Out Loud

For a given node n whose left child is not null, we do the following changes.

n.left.left = n.right;

n.left.right = n;

n.left = null;

n.right = null.

The new root after flipping is the leftmost node in the original tree.

We can solve this recursively bottom up. Solving it top down is more complicated because we need the top nodes to access

the down nodes. Changing the top nodes makes this process a lot harder.

O(n) runtime, O(H) space, H is the max depth of the give binary tree.

 1 public class Solution {
 2     public TreeNode upsideDownBinaryTree(TreeNode root) {
 3         if(root == null || root.left == null) {
 4             return root;
 5         }
 6         TreeNode new_root = upsideDownBinaryTree(root.left);
 7         root.left.left = root.right;
 8         root.left.right = root;
 9         root.left = null;
10         root.right = null;
11         return new_root;
12     }
13 }

Related Problems

Reverse Linked List

[LintCode] Binary Tree Flipping

标签：des toggle linked time example ges hang win dep

原文地址：http://www.cnblogs.com/lz87/p/7478954.html

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