LeetCode | Flatten Binary Tree to Linked List

题目

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

         1
        / \
       2   5
      / \   \
     3   4   6

   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6

需要按先序遍历的方式将树展平。用先序遍历的递归和非递归方法都行，解法1使用的递归的方法。

还有种非递归的写法，没有借助栈就实现了按先序遍历顺序展平，在构造过程中，只要树中有多出来的分叉（左子树），就嫁接到根节点和右子树之间，具体参见解法2。

解法1

public class FlattenBinaryTreeToLinkedList {
	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}

	private TreeNode lastVisit = null;

	public void flatten(TreeNode root) {
		if (root == null) {
			return;
		}
		TreeNode savedRight = root.right;
		if (lastVisit != null) {
			lastVisit.left = null;
			lastVisit.right = root;
		}
		lastVisit = root;
		flatten(root.left);
		flatten(savedRight);
	}
}

public class FlattenBinaryTreeToLinkedList {
	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}

	public void flatten(TreeNode root) {
		while (root != null) {
			if (root.left != null) {
				TreeNode p = root.left;
				while (p.right != null) {
					p = p.right;
				}
				p.right = root.right;
				root.right = root.left;
				root.left = null;
			}
			root = root.right;
		}
	}
}