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
Hints:

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

这里我直接新建了一个tree。代替原来的。注意必须在最后修改root的指针指向的地方，而不是修改root。（值的传递）

 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         // Start typing your Java solution below
13         // DO NOT write main() function
14         if(root == null) return;
15         TreeNode r = new TreeNode(root.val);
16         Stack<TreeNode> s = new Stack<TreeNode>();
17         s.push(root);
18         TreeNode ts = r;
19         while(s.empty() != true){
20             TreeNode cur = s.pop();
21             if(cur.right != null)
22             s.push(cur.right);
23             if(cur.left != null)
24             s.push(cur.left);
25             ts.right = new TreeNode(cur.val);
26             ts = ts.right;
27         }
28         root.right = r.right.right;
29         root.left = null;
30     }
31 }

方法二： 直接修改原来的树。

We can notice that in the flattened tree, each sub node is the successor node of it’s parent node in the pre-order of the original tree. So, we can do it in recursive manner, following the steps below:
1.if root is NULL return;
2.flatten the left sub tree of root, if there is left sub-tree;
3.flatten the right sub-tree of root, if has;
4.if root has no left sub-tree, then root is flattened already, just return;
5.we need to merge the left sub-tree with the right sub-tree, by concatenate the right sub-tree to the last node in left sub-tree.
5.1.find the last node in the left sub tree, as the left is flattened, this is easy.
5.2.concatenate the right sub-tree to this node’s right child.
5.3.move the left sub-tree to the right for root.
5.4.clear the left child of root.
6.done.

 1 A much simpler solution.
 2 public class Solution {
 3 public void flatten(TreeNode root) {
 4 // Start typing your Java solution below
 5 // DO NOT write main() function
 6 preOrder(root, null);
 7 return;
 8 }
 9 
10 public static TreeNode preOrder(TreeNode root, TreeNode prev) {
11 if (root == null) return null;
12 
13 TreeNode leftNode = root.left;
14 TreeNode rightNode = root.right;
15 
16 if (prev == null) {
17 prev = root;
18 } else {
19 prev.left = null;
20 prev.right = root;
21 prev = root;
22 }
23 
24 if (leftNode != null) {
25 prev = preOrder(leftNode, prev);
26 }
27 if (rightNode != null) {
28 prev = preOrder(rightNode, prev);
29 }
30 return prev;
31 }
32 }

 第二遍：

 1 public class Solution {
 2     public void flatten(TreeNode root) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         if(root == null) return;
 6         LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
 7         TreeNode result = null;
 8         queue.push(root);
 9         while(queue.size() != 0){
10             TreeNode tmp = queue.pop();
11             if(tmp.right != null) queue.push(tmp.right);
12             if(tmp.left != null) queue.push(tmp.left);
13             if(result == null) result = tmp;
14             else{
15                 result.right = tmp;
16                 result = result.right;
17             }
18             result.left = result.right = null;
19         }
20     }
21 }

采用stack做preorder search。 存一个current list的tail指针， 更新这个指针。

posted on 2013-09-12 12:48  Step-BY-Step  阅读(222)  评论(0)  编辑  收藏  举报

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