【LeetCode热题100】--114.二叉树展开为链表

114.二叉树展开为链表

方法一：对二叉树进行先序遍历，得到各个节点被访问到的顺序，利用数组存储下来，然后在先序遍历之后更新每个节点的左右节点的信息，将二叉树展开为链表

java 复制代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public void flatten(TreeNode root) {
        List<TreeNode> list = new ArrayList<TreeNode>();
        preorderTraversal(root,list);
        int size = list.size();
        for(int i =  1;i<size;i++){
            TreeNode prev = list.get(i - 1),curr = list.get(i);
            prev.left = null;
            prev.right = curr;
        }
    }
    public void preorderTraversal(TreeNode root, List<TreeNode> list) {
        if (root != null) {
            list.add(root);
            preorderTraversal(root.left, list);
            preorderTraversal(root.right, list);
        }
    }

}

方法二：

java 复制代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public void flatten(TreeNode root) {
        if (root == null) {
            return;
        }
        // 1. 先将左子树展开为链表
        flatten(root.left);
        // 2. 将右子树展开为链表
        flatten(root.right);
        // 将左子树迁移到右子树中
        TreeNode node = root.left;
        if (node != null) {     
            // 如果左子树不为空
            // 3.1. 先找到左子树链表中的最右端的结点
            while (node.right != null) {
                node = node.right;
            }
            // 3.2. 将右子树插入到左子树的尾部结点
            node.right = root.right;
            // 3.3 将左子树换到右结点
            root.right = root.left;
            root.left = null;
        }
    }

}