题解 | #二叉搜索树与双向链表#

这道题主要是:

1.将这些节点从小到大存储到一个数组里面

2.将数组里面的节点进行连接就可以了

/*class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param pRootOfTree TreeNode类 
 * @return TreeNode类
 */
const TreeList:TreeNode[] = []
const inorder = (root:TreeNode)=>{
    if(!root)return
    inorder(root.left)
    TreeList.push(root)
    inorder(root.right)
}
export function Convert(pRootOfTree: TreeNode): TreeNode {
    // 1.将这些节点从小到大存储到一个数组里面
    //2.将数组里面的节点进行连接就可以了
    if(!pRootOfTree)return pRootOfTree
    inorder(pRootOfTree)
    //将节点进行连接
    for(let i = 0;i < TreeList.length - 1;i++){
        TreeList[i].right = TreeList[i+1]
        TreeList[i+1].left = TreeList[i]
    }
    return TreeList[0]
}