Question

Formatted question description: https://leetcode.ca/all/114.html

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)?

Algorithm

DFS

To move the leftmost child node to its parent’s position in a binary tree, we can use the depth-first search (DFS) approach. First, we find the leftmost child node by recursively traversing the left child nodes. Once we reach the leftmost child node, we move back up to its parent node.

To perform the node swap, we disconnect the parent node from its right child node and connect the original left child node to the right child node of the parent node. Then, we connect the original right child node to the right child node of the new right child node. We repeat this operation for each parent node until we have swapped all leftmost child nodes.

Although simple, there are still a few pits

One pitfall to avoid is not properly maintaining a tail node during the traversal. To ensure we add each node to the end of the list, we must save both the left and right child nodes and add them to the tail node.
Additionally, we need to update the tail node every time we add a new node and set the left child node to null, to ensure we only add the right child node to the list.

Flatten process:

Iteration

Starting from the root node, first check whether its left child node exists, if it exists, disconnect the root node from its right child node, and connect the left child node and all the structures behind it to the original right child node. Connect the original right child node to the last right child node of the meta left child node.

Code

import java.util.Stack;

public class Flatten_Binary_Tree_to_Linked_List {

	/**
	 * Definition for a binary tree node.
	 * public class TreeNode {
	 *     int val;
	 *     TreeNode left;
	 *     TreeNode right;
	 *     TreeNode(int x) { val = x; }
	 * }
	 */
    public class Solution {
        public void flatten(TreeNode root) {

            // pre-order traversal
            if (root == null) {
                return;
            }

            Stack<TreeNode> sk = new Stack<>();
            sk.push(root);

            TreeNode prev = new TreeNode(0);

            while (!sk.isEmpty()) {
                TreeNode current = sk.pop();

                if (current.right != null) {
                    sk.push(current.right);
                }
                if (current.left != null) {
                    sk.push(current.left);
                }

                prev.left = null; // cut off
                prev.right = current;

                prev = current;
            }
        }
    }


    public class Solution_recursion {

        TreeNode prev = new TreeNode(0);

        public void flatten(TreeNode root) {

            // pre-order traversal, iteration, then every time stack pop, append to list end
            // maintain the tail node

            if (root == null) {
                return;
            }

            TreeNode left = root.left;
            TreeNode right = root.right;

            // update tail node
            prev.right = root;
            prev.left = null; // @note@note: this is key, cut off left, or "time limit" error
            prev = root;

            if (left != null) flatten(left);
            if (right != null) flatten(right);


        }
    }

}

############

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public void flatten(TreeNode root) {
        while (root != null) {
            if (root.left != null) {
                TreeNode pre = root.left;
                while (pre.right != null) {
                    pre = pre.right;
                }
                pre.right = root.right;
                root.right = root.left;
                root.left = null;
            }
            root = root.right;
        }
    }
}

// OJ: https://leetcode.com/problems/flatten-binary-tree-to-linked-list/
// Time: O(N)
// Space: O(H)
class Solution {
    TreeNode* dfs(TreeNode* root) { // returns a pointer to the last node after flattening
        if (!root || (!root->left && !root->right)) return root;
        auto leftLast = dfs(root->left), rightLast = dfs(root->right), left = root->left, right = root->right;
        root->left = nullptr;
        root->right = left ? left : right;
        if (left) leftLast->right = right;
        return right ? rightLast : leftLast;
    }
public:
    void flatten(TreeNode* root) {
        dfs(root);
    }
};

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution: # no stack
    def flatten(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        while root:
            if root.left:
                pre = root.left
                while pre.right:
                    pre = pre.right
                pre.right = root.right
                root.right = root.left
                root.left = None
            root = root.right


class Solution: # stack, pre-order interation
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if not root:
            return
        
        stack = [root]
        prev = TreeNode(0)
        
        while stack:
            current = stack.pop()
            
            if current.right:
                stack.append(current.right)
            if current.left:
                stack.append(current.left)
                
            prev.left = None
            prev.right = current
            
            prev = current


############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    
    def __init__(self):
        self.prev = TreeNode(0)

    def flatten(self, root: TreeNode) -> None:
        """
        :type root: TreeNode
        :rtype: void Do not return anything, modify root in-place instead.
        """
        if not root:
            return
        
        left = root.left
        right = root.right
        
        self.prev.right = root
        self.prev.left = None # cut
        self.prev = root
        
        if left:
            self.flatten(left)
        if right:
            self.flatten(right)

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

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

/**
 Do not return anything, modify root in-place instead.
 */
function flatten(root: TreeNode | null): void {
    while (root != null) {
        if (root.left != null) {
            let pre = root.left;
            while (pre.right != null) {
                pre = pre.right;
            }
            pre.right = root.right;
            root.right = root.left;
            root.left = null;
        }
        root = root.right;
    }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {void} Do not return anything, modify root in-place instead.
 */
var flatten = function (root) {
    while (root) {
        if (root.left) {
            let pre = root.left;
            while (pre.right) {
                pre = pre.right;
            }
            pre.right = root.right;
            root.right = root.left;
            root.left = null;
        }
        root = root.right;
    }
};

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
// 
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    #[allow(dead_code)]
    pub fn flatten(root: &mut Option<Rc<RefCell<TreeNode>>>) {
        if root.is_none() {
            return;
        }
        let mut v: Vec<Option<Rc<RefCell<TreeNode>>>> = Vec::new();
        // Initialize the vector
        Self::pre_order_traverse(&mut v, root);
        // Traverse the vector
        let n = v.len();
        for i in 0..n - 1 {
            v[i].as_ref().unwrap().borrow_mut().left = None;
            v[i].as_ref().unwrap().borrow_mut().right = v[i + 1].clone();
        }
    }

    #[allow(dead_code)]
    fn pre_order_traverse(v: &mut Vec<Option<Rc<RefCell<TreeNode>>>>, root: &Option<Rc<RefCell<TreeNode>>>) {
        if root.is_none() {
            return;
        }
        v.push(root.clone());
        let left = root.as_ref().unwrap().borrow().left.clone();
        let right = root.as_ref().unwrap().borrow().right.clone();
        Self::pre_order_traverse(v, &left);
        Self::pre_order_traverse(v, &right);
    }
}

114 - Flatten Binary Tree to Linked List

Question

Algorithm

DFS

Iteration

Code

All Problems

All Solutions

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Question

Algorithm

DFS

Iteration

Code

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