【LeetCode】Flatten Binary Tree to Linked List-优快云博客

本文链接：https://blog.youkuaiyun.com/litoupu/article/details/41744969

本文介绍了一种利用先根遍历来找到二叉树最后一个节点，并通过递归方法将二叉树转化为链表的技术。通过实现`getPreOrderLast`和`solve`函数，我们能够有效地对二叉树进行扁平化操作，简化后续的数据处理流程。

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/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {

<span style="white-space:pre">	</span>//获取根节点的先根遍历的最后一个节点
    public TreeNode getPreOrderLast(TreeNode root){
        if(root == null || (root.left == null && root.right == null)) return root;
        TreeNode last = root;
        while(true){
            if(last.right != null) {
                last = last.right;
            }
            else if(last.left != null){
                last = last.left;
            }
            else{
               break;
            }
        }
        return last;
    }

<span style="white-space:pre">	</span>//递归，将二叉树变成一个链表
    public TreeNode solve(TreeNode root){
        if(root == null || (root.left == null && root.right == null)) return root;
        TreeNode rightNode = root.right;
        TreeNode leftNode = root.left;
        TreeNode last = getPreOrderLast(leftNode);
        if(last != null) root.right = solve(leftNode);
        if(last != null) {
            last.right = solve(rightNode); 
            last.left = null;
        }
        if(last == null) root.right = solve(rightNode);//注意，last为空时，root.right 直接为right的链表化结果
        root.left = null;
        return root;
    }
    public void flatten(TreeNode root) {
        if(root == null || (root.left == null && root.right == null))return ;
        solve(root);
    }
}