【LeetCode热题100】--114.二叉树展开为链表-CSDN博客

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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

/**
 * 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

/**
 * 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;
        }
    }

}