新手学习算法----二叉树（将一个二叉查找树按照中序遍历转换成双向链表）

Posted 2020-09-29

题目：将一个二叉查找树按照中序遍历转换成双向链表。

         

给定一个二叉查找树：

    4
   /   2   5
 / 1   3

返回 1<->2<->3<->4<->5。

思路：如果对于当前节点，把右子树转换成双向链表，然后把左子树转换成双向链表，转换的时候我们都标记了链表的头节点和尾节点，那么只需要将当前节点和左子树的尾部相连，和右子树的头部相连即可。

Java代码：这个是借鉴九章里面的解题法。但是对于左右子树转换成二叉树也不是很理解，还待需要继续分析。

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Definition for Doubly-ListNode.
 * public class DoublyListNode {
 *     int val;
 *     DoublyListNode next, prev;
 *     DoublyListNode(int val) {
 *         this.val = val;
 *         this.next = this.prev = null;
 *     }
 * }
 */ 
 
class ResultType {
    DoublyListNode first, last;
    
    public ResultType(DoublyListNode first, DoublyListNode last) {
        this.first = first;
        this.last = last;
    }
}

public class Solution {
    /**
     * @param root: The root of tree
     * @return: the head of doubly list node
     */
    public DoublyListNode bstToDoublyList(TreeNode root) {  
        if (root == null) {
            return null;
        }
        
        ResultType result = helper(root);
        return result.first;
    }
    
    public ResultType helper(TreeNode root) {  
        if (root == null) {
            return null;
        }
        
        ResultType left = helper(root.left);
        ResultType right = helper(root.right);
        DoublyListNode node = new DoublyListNode(root.val);
        
        ResultType result = new ResultType(null, null);
        
        if (left == null) {
            result.first = node;
        } else {
            result.first = left.first;//?????
            left.last.next = node;
            node.prev = left.last;
        }
        
        if (right == null) {
            result.last = node;
        } else {
            result.last = right.last;//?????
            right.first.prev = node;
            node.next = right.first;
        }
        
        return result;
    }
}