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Leetcode: 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   6The flattened tree should look like:   1         2             3                 4                     5                         6click 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.

难度:70

把先序遍历的结果存到一个ArrayList<TreeNode>里面,然后从根节点开始,依次把左子树置空,右子树置为ArrayList里面存的下一个节点

 1 /** 2  * Definition for binary tree 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     public void flatten(TreeNode root) {12         if (root == null) return;13         ArrayList<TreeNode> preorder = new ArrayList<TreeNode>();14         helper(root, preorder);15         preorder.remove(0);16         TreeNode temp = root;17         while (preorder.size()!=0) {18             temp.left = null;19             temp.right = preorder.get(0);20             preorder.remove(0);21             temp = temp.right;22         }23     }24     25     public void helper(TreeNode root, ArrayList<TreeNode> preorder) {26         if (root == null) return;27         preorder.add(root);28         helper(root.left, preorder);29         helper(root.right, preorder);30     }31 }

需要注意看一个arraylist是否为减为空应该是看它的size是否等于0,而不是arraylist == null,后者表示没有给该arraylist分配地址

Leetcode: Flatten Binary Tree to Linked List