LeetCode刷题笔录Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

         1
        / \
       2   5
      / \   \
     3   4   6

The flattened tree should look like:

   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6

click to show hints.

Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

按照Hint所说的，用一个preorder traversal就搞定了。previous记录前一个TreeNode。这里我用的是非递归的方法。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void flatten(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        if(root == null)
            return ;
        stack.push(root);
        TreeNode previous = null;
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            if(previous == null)
                previous = node;
            else{
                previous.right = node;
                previous.left = null;
                previous = node;
            }
            
            if(node.right != null)
                stack.push(node.right);
            if(node.left != null)
                stack.push(node.left);
        }
        
        return;
    }
}

不用递归还是因为对于递归用的不熟练。

递归的方法在这里有。

当然这个解法是用了extra space的，所以并不是题目要求的解法。上面那个连接里还有一个非递归不用stack的方法，直接参考那里的解法好了。