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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 next pointer in ListNode.
1
/ 2 5
/ \ 3 4 6
The flattened tree should look like:
1
2
3
4
5
6
分析：
把所有的树节点按照PRE-ORDER的顺序存在ArrayList里面，然后遍历ArrayList里的节点，调整它们的left and right child指针。
1 /** 2 * Definition of TreeNode: 3 * public class TreeNode { 4 * public int val; 5 * public TreeNode left, right; 6 * public TreeNode(int val) { 7 * this.val = val; 8 * this.left = this.right = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 /** 14 * @param root: a TreeNode, the root of the binary tree 15 * @return: nothing 16 * cnblogs.com/beiyeqingteng 17 */ 18 public void flatten(TreeNode root) { 19 // write your code here 20 ArrayList<TreeNode> list = new ArrayList<TreeNode>(); 21 preOrder(root, list); 22 23 for (int i = 0; i < list.size(); i++) { 24 list.get(i).left = null; 25 if (i == list.size() - 1) { 26 list.get(i).right = null; 27 } else { 28 list.get(i).right = list.get(i + 1); 29 } 30 } 31 } 32 33 public void preOrder(TreeNode root, ArrayList<TreeNode> list) { 34 if (root != null) { 35 list.add(root); 36 preOrder(root.left, list); 37 preOrder(root.right, list); 38 } 39 } 40 }
还有一种是递归的方式，下面的方法参考了另一个网友的做法，觉得这种方法非常容易理解，而且具有一般性。
解法2：递归构建
假设某节点的左右子树T(root->left)和T(root->right)已经flatten成linked list了：
1
/ \
2 5
\ \
3 6 <- rightTail
\
4 <- leftTail
如何将root、T(root->left)、T(root->right) flatten成一整个linked list？显而易见：
temp = root->right
root->right = root->left
root->left = NULL
leftTail->right = temp
这里需要用到已经flatten的linked list的尾部元素leftTail, rightTail。所以递归返回值应该是生成的整个linked list的尾部。
1 class Solution { 2 public void flatten(TreeNode root) { 3 flattenBT(root); 4 } 5 6 TreeNode flattenBT(TreeNode root) { 7 if(root == null) return null; 8 TreeNode leftTail = flattenBT(root.left); 9 TreeNode rightTail = flattenBT(root.right); 10 if(root.left != null) { 11 TreeNode temp = root.right; 12 root.right = root.left; 13 root.left = null; 14 leftTail.right = temp; 15 } 16 17 if(rightTail != null) return rightTail; 18 if(leftTail != null) return leftTail; 19 return root; 20 } 21 }
需注意几个细节
ln 10：只有当左子树存在时才将它插入右子树中
ln 17-19：返回尾部元素时，需要特殊处理 (1) 有右子树的情况，(2) 无右子树但有左子树的情况，(3)左右子树均不存在的情况。
Reference:
http://bangbingsyb.blogspot.com/2014/11/leetcode-flatten-binary-tree-to-linked.html
Flatten Binary Tree to Linked List
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5631722.html
