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
思路:这题主要是就是前序遍历。主要解法就是将左子树转换为右支树,同一时候加入在右子树前。左子树求解时,须要主要左子树的深度。
详细代码例如以下:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public void flatten(TreeNode root) { if(root == null){ return; } TreeNode r = new TreeNode(0); fun(root,r); root = r.right; } private boolean fun(TreeNode root, TreeNode t){ if(root == null){ return false; } TreeNode p = root.left; TreeNode q = root.right; t.right = root; t.left = null; if(p != null){ if(fun(p, t.right)){ if(q != null){ TreeNode k = t.right; while(k.right != null){ k = k.right; } fun(q,k); } }else{ if(q != null){ fun(q,t.right); } } }else{ if(q != null){ fun(q,t.right); } } return true; } }