114. Flatten Binary Tree to Linked List

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SnailRunFast

最新推荐文章于 2024-08-25 16:11:27 发布

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本文介绍了一种将二叉树结构转换为链表的方法，通过递归或栈辅助，确保转换后的链表中节点按原二叉树前序遍历顺序排列。讨论了两种实现方式，一种使用栈迭代，另一种采用递归调用。

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114. Flatten Binary Tree to Linked List

Medium

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

For example, given the following tree:

The flattened tree should look like:

/**
 * 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) {
        if (null != root) {
            Stack<TreeNode> stack = new Stack<TreeNode>();
            stack.push(root);
            while (!stack.isEmpty()) {
                TreeNode node = stack.pop();

                if (node.right != null) {
                    stack.push(node.right);
                }
                if (node.left != null) {
                    stack.push(node.left);
                }
                
                node.right = stack.isEmpty() ? node.right : stack.peek();
                node.left = null;
            } 
        }

    }
}

/**
 * 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) {
        innerFlatten(root);
    }
    
    private TreeNode innerFlatten(TreeNode root) {
        if (null == root) {
            return root;
        }
        
        TreeNode left = innerFlatten(root.left);
        TreeNode right = innerFlatten(root.right);
        if (left != null && right != null) {
            TreeNode mostRight = left;
            while (mostRight.right != null) {
                mostRight = mostRight.right;
            }
            mostRight.right = right;
            root.right = left;
            root.left = null;
            
        } else if (left != null && null == right) {
            root.right = left;
            root.left = null;
        }
        return root;
    }
}