LeetCode之flatten binary tree

问题描述：

/** * 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 */

把一个二叉树flatten 成一个像链表一样的树，如上图所示。
可以看出，结果树就是二叉树前序遍历的结果顺序。
所以，可以先把二叉树前序遍历一下，把遍历的结果存储到队列queue当中，再构建新的结果二叉树即可。
代码如下：

public void flatten(TreeNode root) {    if (root == null) return;    Stack<TreeNode> stack = new Stack<TreeNode>();    Queue<TreeNode> queue = new LinkedList<TreeNode>();    stack.push(root);    queue.add(root);    while (!stack.isEmpty()) {      TreeNode top = stack.peek();      if (top.left != null) {        queue.add(top.left);        stack.push(top.left);        top.left = null;      } else if (top.right != null) {          queue.add(top.right);          stack.push(top.right);          top.right = null;      } else { //top.left == null && top.right == null        stack.pop();      }    }    TreeNode node = queue.poll();    while (!queue.isEmpty()) {      node.right = queue.poll();      node = node.right;    }  }

上面这种方法借用了队列queue和栈stack。下面这种方法也很巧妙：

public void flatten(TreeNode root) {        if (root == null) return;        if (root.left != null) {            TreeNode rightMost = findRightMost(root.left);            TreeNode rightChild = root.right;            root.right = root.left;            root.left = null;            rightMost.right = rightChild;        }        flatten(root.right);    }    private TreeNode findRightMost(TreeNode root) {        if (root == null || root.right == null) return root;        return findRightMost(root.right);    }