binary-tree-tilt

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Algorithms

Given the <code>root</code> of a binary tree, return the sum of every tree node&#39;s tilt.

The tilt of a tree node is the absolute difference between the sum of all left subtree node values and all right subtree node values. If a node does not have a left child, then the sum of the left subtree node values is treated as <code>0</code>. The rule is similar if the node does not have a right child.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/20/tilt1.jpg" style="width: 712px; height: 182px;" />
<pre>
Input: root = [1,2,3]
Output: 1
Explanation: 
Tilt of node 2 : |0-0| = 0 (no children)
Tilt of node 3 : |0-0| = 0 (no children)
Tilt 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)
Sum of every tilt : 0 + 0 + 1 = 1
</pre>

Example 2:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/20/tilt2.jpg" style="width: 800px; height: 203px;" />
<pre>
Input: root = [4,2,9,3,5,null,7]
Output: 15
Explanation: 
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 5 : |0-0| = 0 (no children)
Tilt of node 7 : |0-0| = 0 (no children)
Tilt 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)
Tilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7)
Tilt 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)
Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15
</pre>

Example 3:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/20/tilt3.jpg" style="width: 800px; height: 293px;" />
<pre>
Input: root = [21,7,14,1,1,2,2,3,3]
Output: 9
</pre>

&nbsp;
Constraints:

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 104]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
</ul>

findTilt

Binary Tree Tilt

Find All The Lonely Nodes

find-all-the-lonely-nodes

寻找所有的独生节点

Tree

Depth-First Search

Binary Tree

给你一个二叉树的根节点 <code>root</code> ，计算并返回 整个树 的坡度 。

一个树的 节点的坡度 定义即为，该节点左子树的节点之和和右子树节点之和的 差的绝对值 。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 。

整个树 的坡度就是其所有节点的坡度之和。

&nbsp;

示例 1：
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/20/tilt1.jpg" style="width: 712px; height: 182px;" />
<pre>
输入：root = [1,2,3]
输出：1
解释：
节点 2 的坡度：|0-0| = 0（没有子节点）
节点 3 的坡度：|0-0| = 0（没有子节点）
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 ）
坡度总和：0 + 0 + 1 = 1
</pre>

示例 2：
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/20/tilt2.jpg" style="width: 800px; height: 203px;" />
<pre>
输入：root = [4,2,9,3,5,null,7]
输出：15
解释：
节点 3 的坡度：|0-0| = 0（没有子节点）
节点 5 的坡度：|0-0| = 0（没有子节点）
节点 7 的坡度：|0-0| = 0（没有子节点）
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 ）
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 ）
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 ）
坡度总和：0 + 0 + 0 + 2 + 7 + 6 = 15
</pre>

示例 3：
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/20/tilt3.jpg" style="width: 800px; height: 293px;" />
<pre>
输入：root = [21,7,14,1,1,2,2,3,3]
输出：9
</pre>

&nbsp;

提示：

<ul>
	<li>树中节点数目的范围在 <code>[0, 104]</code> 内</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
</ul>