897. Increasing Order Search Tree.md

897. Increasing Order Search Tree

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leetcode Daily Challenge on December 3rd, 2020.

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Difficulty : Easy

Related Topics : DFS、Tree

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Given a binary search 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

Constraints:

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

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Solution

• java

  • mine

    DFS Runtime: 0 ms, faster than 100.00%, Memory Usage: 37.4 MB, less than 100.00% of Java online submissions

    public TreeNode increasingBST(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        dfs(root, list);
        Integer[] arr = new Integer[list.size()];
        Arrays.sort(list.toArray(arr));
        TreeNode res = new TreeNode(arr[0]);
        TreeNode temp = res;
        int index = 1;
        while(index < list.size()){
            TreeNode node = new TreeNode(arr[index]);
            temp.right = node;
            temp = node;
            index++;
        }
        return res;
    }
    public void dfs(TreeNode node, List<Integer> list){
        if(node == null){
            return;
        }
        list.add(node.val);
        dfs(node.left,list);
        dfs(node.right,list);
    }
    

  • the most votes Runtime: 0 ms, faster than 100.00%，Memory Usage: 36.9 MB, less than 100.00% of Java online submissions

    //O(N)Time O(height)Space
    public TreeNode increasingBST(TreeNode root) {
        return increasingBST(root, null);
    }
    public TreeNode increasingBST(TreeNode root, TreeNode tail) {
        if (root == null) return tail;
        TreeNode res = increasingBST(root.left, root);
        root.left = null;
        root.right = increasingBST(root.right, tail);
        return res;
    }