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Flatten a binary tree to a fake "linked list" in pre-order traversal.
Here we use the right pointer in TreeNode as the nextpointer in ListNode.
1
1 2
/ \ 2 5 => 3
/ \ \ 3 4 6 4
5
6
Don‘t forget to mark the left child of each node to null. Or you will get Time Limit Exceeded or Memory Limit Exceeded.
Do it in-place without any extra memory.
递归一下，注意如果左子树为null直接flatten右子树然后返回，否则超时
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: a TreeNode, the root of the binary tree * @return: nothing */ public void flatten(TreeNode root) { // write your code here if(root == null) return; if(root.left == null && root.right == null) return; if(root.left != null){ flatten(root.left); } if(root.right != null){ flatten(root.right); } TreeNode temp = root.right; if(root.left != null){ root.right = root.left; root.left = null; TreeNode p = root; while(p.right != null) p = p.right; p.right = temp; } return; } }
lintcode-easy-Flatten Binary Tree to Linked List
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原文地址：http://www.cnblogs.com/goblinengineer/p/5213925.html
