modify Archives - Huahua's Tech Road https://zxi.mytechroad.com/blog/tag/modify/ Wed, 12 Feb 2020 06:50:17 +0000 en hourly 1 https://wordpress.org/?v=6.0.8 https://zxi.mytechroad.com/blog/wp-content/uploads/2017/09/cropped-photo-32x32.jpg modify Archives - Huahua's Tech Road https://zxi.mytechroad.com/blog/tag/modify/ 32 32 花花酱 LeetCode 897. Increasing Order Search Tree https://zxi.mytechroad.com/blog/tree/leetcode-897-increasing-order-search-tree/ https://zxi.mytechroad.com/blog/tree/leetcode-897-increasing-order-search-tree/#respond Sun, 02 Sep 2018 05:42:34 +0000 https://zxi.mytechroad.com/blog/?p=3801 Problem Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node…

The post 花花酱 LeetCode 897. Increasing Order Search Tree appeared first on Huahua's Tech Road.

]]> Problem

Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.

Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

       5
      / \
    3    6
   / \    \
  2   4    8
 /        / \ 
1        7   9

Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

 1
  \
   2
    \
     3
      \
       4
        \
         5
          \
           6
            \
             7
              \
               8
                \
                 9

Note:

  1. The number of nodes in the given tree will be between 1 and 100.
  2. Each node will have a unique integer value from 0 to 1000.

Solution: In-order traversal

root = 5
inorder(root.left) 之后
self.prev = 4
（1-4）已经处理完了，这时候的树是很奇怪的一个形状，3即是2的右子树，又是5的左子树。
   1
    \
     2    5
      \  /  \ 
       3     6
        \     \
 prev -> 4     8
             /  \
            7    9
—————————
5.left = None # 把5->3的链接断开
5
 \
  6
   \
    8
   /  \
  7    9
—————————–
self.prev.right = root  <=> 4.right = 5
把5接到4的右子树
1
 \
  2
   \
    3
     \
      4
       \
        5 <– prev
         \
          6
           \
            8
          /   \
         7     9
self.prev = root <=> prev = 5
inorder(5.right) <=> inorder(6) 然后再去递归处理6（及其子树）即可。

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua
// Running time: 64 ms
class Solution {
public:
  TreeNode* increasingBST(TreeNode* root) {
    TreeNode dummy(0);
    prev_ = &dummy;
    inorder(root);
    return dummy.right;
  }
private:  
  TreeNode* prev_;
  void inorder(TreeNode* root) {
    if (root == nullptr) return;
    inorder(root->left);
    prev_->right = root;  
    prev_ = root;
    prev_->left = nullptr;
    inorder(root->right);
  }
};

Java

class Solution {
  private TreeNode prev;
  
  public TreeNode increasingBST(TreeNode root) {
    TreeNode dummy = new TreeNode(0);
    this.prev = dummy;
    inorder(root);
    return dummy.right;
  }
  
  private void inorder(TreeNode root) {
    if (root == null) return;
    inorder(root.left);
    this.prev.right = root;
    this.prev = root;
    this.prev.left = null;
    inorder(root.right);
  }
}

Python3

class Solution:
  def increasingBST(self, root):
    dummy = TreeNode(0)
    self.prev = dummy
    def inorder(root):
      if not root: return None
      inorder(root.left)
      root.left = None
      self.prev.right = root
      self.prev = root
      inorder(root.right)
    inorder(root)
    return dummy.right

The post 花花酱 LeetCode 897. Increasing Order Search Tree appeared first on Huahua's Tech Road.

]]> https://zxi.mytechroad.com/blog/tree/leetcode-897-increasing-order-search-tree/feed/ 0