mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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183 lines
27 KiB
JSON
183 lines
27 KiB
JSON
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"categoryTitle": "Algorithms",
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"boundTopicId": 1305,
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"title": "Binary Tree Tilt",
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"titleSlug": "binary-tree-tilt",
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"content": "<p>Given the <code>root</code> of a binary tree, return <em>the sum of every tree node's <strong>tilt</strong>.</em></p>\n\n<p>The <strong>tilt</strong> of a tree node is the <strong>absolute difference</strong> between the sum of all left subtree node <strong>values</strong> and all right subtree node <strong>values</strong>. If a node does not have a left child, then the sum of the left subtree node <strong>values</strong> is treated as <code>0</code>. The rule is similar if the node does not have a right child.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/20/tilt1.jpg\" style=\"width: 712px; height: 182px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,3]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> \nTilt of node 2 : |0-0| = 0 (no children)\nTilt of node 3 : |0-0| = 0 (no children)\nTilt of node 1 : |2-3| = 1 (left subtree is just left child, so sum is 2; right subtree is just right child, so sum is 3)\nSum of every tilt : 0 + 0 + 1 = 1\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/20/tilt2.jpg\" style=\"width: 800px; height: 203px;\" />\n<pre>\n<strong>Input:</strong> root = [4,2,9,3,5,null,7]\n<strong>Output:</strong> 15\n<strong>Explanation:</strong> \nTilt of node 3 : |0-0| = 0 (no children)\nTilt of node 5 : |0-0| = 0 (no children)\nTilt of node 7 : |0-0| = 0 (no children)\nTilt of node 2 : |3-5| = 2 (left subtree is just left child, so sum is 3; right subtree is just right child, so sum is 5)\nTilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7)\nTilt of node 4 : |(3+5+2)-(9+7)| = |10-16| = 6 (left subtree values are 3, 5, and 2, which sums to 10; right subtree values are 9 and 7, which sums to 16)\nSum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15\n</pre>\n\n<p><strong>Example 3:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/20/tilt3.jpg\" style=\"width: 800px; height: 293px;\" />\n<pre>\n<strong>Input:</strong> root = [21,7,14,1,1,2,2,3,3]\n<strong>Output:</strong> 9\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.</li>\n\t<li><code>-1000 <= Node.val <= 1000</code></li>\n</ul>\n",
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"translatedTitle": "二叉树的坡度",
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"translatedContent": "<p>给你一个二叉树的根节点 <code>root</code> ，计算并返回 <strong>整个树 </strong>的坡度 。</p>\n\n<p>一个树的<strong> 节点的坡度 </strong>定义即为，该节点左子树的节点之和和右子树节点之和的 <strong>差的绝对值 </strong>。如果没有左子树的话，左子树的节点之和为 0 ；没有右子树的话也是一样。空结点的坡度是 0 。</p>\n\n<p><strong>整个树</strong> 的坡度就是其所有节点的坡度之和。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/20/tilt1.jpg\" style=\"width: 712px; height: 182px;\" />\n<pre>\n<strong>输入：</strong>root = [1,2,3]\n<strong>输出：</strong>1\n<strong>解释：</strong>\n节点 2 的坡度：|0-0| = 0（没有子节点）\n节点 3 的坡度：|0-0| = 0（没有子节点）\n节点 1 的坡度：|2-3| = 1（左子树就是左子节点，所以和是 2 ；右子树就是右子节点，所以和是 3 ）\n坡度总和：0 + 0 + 1 = 1\n</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/20/tilt2.jpg\" style=\"width: 800px; height: 203px;\" />\n<pre>\n<strong>输入：</strong>root = [4,2,9,3,5,null,7]\n<strong>输出：</strong>15\n<strong>解释：</strong>\n节点 3 的坡度：|0-0| = 0（没有子节点）\n节点 5 的坡度：|0-0| = 0（没有子节点）\n节点 7 的坡度：|0-0| = 0（没有子节点）\n节点 2 的坡度：|3-5| = 2（左子树就是左子节点，所以和是 3 ；右子树就是右子节点，所以和是 5 ）\n节点 9 的坡度：|0-7| = 7（没有左子树，所以和是 0 ；右子树正好是右子节点，所以和是 7 ）\n节点 4 的坡度：|(3+5+2)-(9+7)| = |10-16| = 6（左子树值为 3、5 和 2 ，和是 10 ；右子树值为 9 和 7 ，和是 16 ）\n坡度总和：0 + 0 + 0 + 2 + 7 + 6 = 15\n</pre>\n\n<p><strong>示例 3：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/20/tilt3.jpg\" style=\"width: 800px; height: 293px;\" />\n<pre>\n<strong>输入：</strong>root = [21,7,14,1,1,2,2,3,3]\n<strong>输出：</strong>9\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中节点数目的范围在 <code>[0, 10<sup>4</sup>]</code> 内</li>\n\t<li><code>-1000 <= Node.val <= 1000</code></li>\n</ul>\n",
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"code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n * public var val: Int\n * public var left: TreeNode?\n * public var right: TreeNode?\n * public init() { self.val = 0; self.left = nil; self.right = nil; }\n * public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }\n * public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {\n * self.val = val\n * self.left = left\n * self.right = right\n * }\n * }\n */\nclass Solution {\n func findTilt(_ root: TreeNode?) -> Int {\n\n }\n}",
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"code": "/**\n * Example:\n * var ti = TreeNode(5)\n * var v = ti.`val`\n * Definition for a binary tree node.\n * class TreeNode(var `val`: Int) {\n * var left: TreeNode? = null\n * var right: TreeNode? = null\n * }\n */\nclass Solution {\n fun findTilt(root: TreeNode?): Int {\n\n }\n}",
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"code": "// Definition for a binary tree node.\n// #[derive(Debug, PartialEq, Eq)]\n// pub struct TreeNode {\n// pub val: i32,\n// pub left: Option<Rc<RefCell<TreeNode>>>,\n// pub right: Option<Rc<RefCell<TreeNode>>>,\n// }\n//\n// impl TreeNode {\n// #[inline]\n// pub fn new(val: i32) -> Self {\n// TreeNode {\n// val,\n// left: None,\n// right: None\n// }\n// }\n// }\nuse std::rc::Rc;\nuse std::cell::RefCell;\nimpl Solution {\n pub fn find_tilt(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {\n\n }\n}",
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"Don't think too much, this is an easy problem. Take some small tree as an example.",
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"Can a parent node use the values of its child nodes? How will you implement it?",
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"May be recursion and tree traversal can help you in implementing.",
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