Leetcode - Increasing Order Search Tree

https://leetcode.com/problems/increasing-order-search-tree/

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  

func increasingBST(root *TreeNode) *TreeNode {
    result := []int{}
    stack := []*TreeNode{}
    node := root
    for node != nil || len(stack) != 0 {
        for node != nil {
            stack = append(stack, node)
            node = node.Left
        }
        if len(stack) != 0 {
            node = stack[len(stack)-1]
            result = append(result, node.Val)
            stack = stack[:len(stack)-1]
            node = node.Right
        }
    }
    //fmt.Println("result: ", result)
    root = &TreeNode{}
    node = root
    for i := 0; i < len(result); i++ {
        node.Right = &TreeNode{
            Val: result[i],
        }
        node = node.Right
    }
    return root.Right
}

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