每天1题算法题（5）- 二叉树展开为链表

给定一个二叉树，原地将它展开为一个单链表。

 

 解答

将二叉树展开为单链表之后，单链表中的节点顺序即为二叉树的前序遍历访问各节点的顺序

递归实现

class Solution {

    private TreeNode curNode;

    public void flatten(TreeNode root) {
        List<TreeNode> treeNodeList = new ArrayList();
        preOrder(root, treeNodeList);
        for(int i=1; i < treeNodeList.size(); i++) {
            root.right = treeNodeList.get(i);
            root.left = null;
            root = root.right;
        }
       
    }

    private void preOrder(TreeNode node, List<TreeNode> treeNodeList) {
        if(node == null) {
            return;
        }
        treeNodeList.add(node);
        preOrder(node.left, treeNodeList);
        preOrder(node.right, treeNodeList);
    }
}

 迭代实现

class Solution {

    private TreeNode curNode;

    public void flatten(TreeNode root) {
        if(root == null) {
            return;
        }
        List<TreeNode> list = new ArrayList();
        Stack<TreeNode> stack = new Stack();
        stack.push(root);
        while(!stack.isEmpty()) {
            TreeNode treeNode = stack.pop();
            list.add(treeNode);

            if(treeNode.right != null){
                stack.push(treeNode.right);
            }
            if(treeNode.left != null){
                stack.push(treeNode.left);
            }
            
        }

        for(int i=1;i<list.size();i++) {
            root.right = list.get(i);
            root.left = null;
            root = root.right;
        }
    }
}