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给定一个二叉树的根 root 和两个整数 val 和 depth ,在给定的深度 depth 处添加一个值为 val 的节点行。
注意,根节点 root 位于深度 1 。
加法规则如下:
输入: root = [4,2,6,3,1,5], val = 1, depth = 2 输出: [4,1,1,2,null,null,6,3,1,5]
输入: root = [4,2,null,3,1], val = 1, depth = 3 输出: [4,2,null,1,1,3,null,null,1]
来源:力扣(LeetCode) 链接:https://leetcode.cn/problems/add-one-row-to-tree 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
The text was updated successfully, but these errors were encountered:
比较简单的树遍历,可以借助 php 的匿名函数递归 yankewei/blog#8
/** * Definition for a binary tree node. * class TreeNode { * public $val = null; * public $left = null; * public $right = null; * function __construct($val = 0, $left = null, $right = null) { * $this->val = $val; * $this->left = $left; * $this->right = $right; * } * } */ class Solution { /** * @param TreeNode $root * @param Integer $val * @param Integer $depth * @return TreeNode */ function addOneRow($root, $val, $depth) { if ($depth === 1) { return new TreeNode($val, $root, null); } $dfs = function (?TreeNode $root, int $level) use (&$dfs, $val, $depth){ if ($level === $depth-1) { $root->left = new TreeNode($val, $root->left, null); $root->right = new TreeNode($val, null, $root->right); return; } if ($root->left !== null) { $dfs($root->left, $level + 1); } if ($root->right !== null) { $dfs($root->right, $level + 1); } }; $dfs($root, 1); // level 设置为1,为了获取正确的节点 return $root; } }
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给定一个二叉树的根 root 和两个整数 val 和 depth ,在给定的深度 depth 处添加一个值为 val 的节点行。
注意,根节点 root 位于深度 1 。
加法规则如下:
示例 1:
示例 2:
提示:
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/add-one-row-to-tree
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
The text was updated successfully, but these errors were encountered: