20220201.md

Algorithm

114. Flatten Binary Tree to Linked List

Description

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

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

private TreeNode prev = null;

public void flatten(TreeNode root) {
    if (root == null)
        return;
    flatten(root.right);
    flatten(root.left);
    root.right = prev;
    root.left = null;
    prev = root;
}

Discuss

1
/ \
2   5
/ \   \
3   4   6
-----------        
pre = 5
cur = 4

1
/
2   
/ \   
3   4
 \
  5
   \
    6
-----------        
pre = 4
cur = 3

1
/
2   
/   
3
\
4
\
5
 \
  6
-----------        
cur = 2
pre = 3

1
/
2   
\
3
 \
  4
   \
    5
     \
      6
-----------        
cur = 1
pre = 2

1
\
2
\
3
 \
  4
   \
    5
     \
      6

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