【LeetCode热题100（45/100）】二叉树展开为链表

题目地址：链接

思路： 每次遍历时，获取左右节点展开为链表的内容，将左节点接到父亲对右节点，原父节点接到上述左节点的最右节点

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {void} Do not return anything, modify root in-place instead.
 */
var flatten = function(root) {
    const dfs = (root) => {
        if(!root) return null;
        let p = root;
        let [left, right] = [root.left, root.right];
        left = dfs(left);
        right= dfs(right);
        
        p.left = null;
        p.right = left;
        
        while(p.right) p = p.right;
        p = right;

        return root;
    }
    dfs(root);
};