LeetCode-in-All

114. Flatten Binary Tree to Linked List

Medium

Given the root of a binary tree, flatten the tree into a “linked list”:

• The “linked list” should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
• The “linked list” should be in the same order as a pre-order traversal of the binary tree.

Example 1:

Input: root = [1,2,5,3,4,null,6]

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

Example 2:

Input: root = []

Output: []

Example 3:

Input: root = [0]

Output: [0]

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -100 <= Node.val <= 100

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

Solution

type TreeNode struct {
	Val   int
	Left  *TreeNode
	Right *TreeNode
}

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func flatten(root *TreeNode) {
	if root == nil {
		return
	}
	if root.Left != nil {
		flatten(root.Left)
	}
	if root.Right != nil {
		flatten(root.Right)
	}
	if root.Left != nil {
		rootRight := root.Right
		root.Right = root.Left
		t := root.Left
		for t.Right != nil {
			t = t.Right
		}
		t.Right = rootRight
	}
	root.Left = nil
}

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