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Merged

pull merged 1 commit into AlgorithmAndLeetCode:master from labuladong:master

May 9, 2023

+10 −0

Merged

[pull] master from labuladong:master #89

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labuladong May 9, 2023

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6 changes: 6 additions & 0 deletions 数据结构系列/二叉树总结.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -282,6 +282,8 @@ void traverse(TreeNode root) {
		}
		```

	<visual slug='mydata-maxdepth1' title='遍历思路的可视化' />

		这个解法应该很好理解,但为什么需要在前序位置增加 `depth`,在后序位置减小 `depth`?

		因为前面说了,前序位置是进入一个节点的时候,后序位置是离开一个节点的时候,`depth` 记录当前递归到的节点深度,你把 `traverse` 理解成在二叉树上游走的一个指针,所以当然要这样维护。
Expand Down Expand Up		@@ -310,6 +312,8 @@ int maxDepth(TreeNode root) {
		}
		```

	<visual slug='mydata-maxdepth2' title='分解问题思路的可视化' />

		只要明确递归函数的定义,这个解法也不难理解,但为什么主要的代码逻辑集中在后序位置?

		因为这个思路正确的核心在于,你确实可以通过子树的最大深度推导出原树的深度,所以当然要首先利用递归函数的定义算出左右子树的最大深度,然后推出原树的最大深度,主要逻辑自然放在后序位置。
Expand Down Expand Up		@@ -548,6 +552,8 @@ class Solution {
		}
		```

	<visual slug='mydata-diameter-of-binary-tree'/>

		这下时间复杂度只有 `maxDepth` 函数的 O(N) 了。

		讲到这里,照应一下前文:遇到子树问题,首先想到的是给函数设置返回值,然后在后序位置做文章。
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2 changes: 2 additions & 0 deletions 数据结构系列/二叉树系列1.md

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Expand Up		@@ -341,6 +341,8 @@ void flatten(TreeNode root) {
		}
		```

	<visual slug='flatten-binary-tree-to-linked-list' />

		你看,这就是递归的魅力,你说 `flatten` 函数是怎么把左右子树拉平的?

		不容易说清楚,但是只要知道 `flatten` 的定义如此并利用这个定义,让每一个节点做它该做的事情,然后 `flatten` 函数就会按照定义工作。
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2 changes: 2 additions & 0 deletions 算法思维系列/双指针技巧.md

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Expand Up		@@ -93,6 +93,8 @@ int removeDuplicates(int[] nums) {
		}
		```

	<visual slug='remove-duplicates-from-sorted-array' />

		算法执行的过程如下 GIF 图:

		![](https://labuladong.github.io/pictures/数组去重/1.gif)
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