本文共 1850 字，大约阅读时间需要 6 分钟。

方法一 哈希表

/** * 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 {
       List
   
     list = new ArrayList<>();    public void flatten(TreeNode root) {
                   help(root);        TreeNode temp = new TreeNode(-1);        for(int i = 0; i < list.size(); i++) {
               temp.right = list.get(i);            temp = temp.right;            temp.left = null;        }    }    public void help(TreeNode root) {
           if(root == null) {
               return;        }        list.add(root);        help(root.left);        help(root.right);    }}
   

方法二 迭代法

/** * 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;        }        TreeNode pre = null;        Stack
   
     stack = new Stack<>();        stack.push(root);        while(!stack.isEmpty()) {
               TreeNode cur = stack.pop();            if(pre != null) {
   //用来保存前一个节点                pre.left = null;                pre.right = cur;            }            if(cur.right != null) {
                   stack.push(cur.right);            }            if(cur.left != null) {
                   stack.push(cur.left);            }            pre = cur;        }    }}
   

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