题目描述
给定一个二叉树,原地将它展开为链表。
例如,给定二叉树
1
/ \
2 5
/ \ \
3 4 6
将其展开为:
1
\
2
\
3
\
4
\
5
\
6
思路
大体思路就是左子树移到右子树的位置,右子树接在之后。每一棵子树也要同样展开,就是递归。
解法
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {void} Do not return anything, modify root in-place instead.
*/
var flatten = function(root) {
if (!root || (!root.left && !root.right)) return root;
const right = root.right;
root.right = flatten(root.left);
root.left = null;
let tail = root;
while (tail.right) {
tail = tail.right;
}
tail.right = flatten(right);
return root;
};