一刷1104

Author: diguage

README.adoc

@@ -7753,14 +7753,14 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|{doc_base_url}/1103-distribute-candies-to-people.adoc[题解]
 //|Easy
 //|
-//
-//|{counter:codes}
-//|{leetcode_base_url}/path-in-zigzag-labelled-binary-tree/[1104. Path In Zigzag Labelled Binary Tree^]
-//|{source_base_url}/_1104_PathInZigzagLabelledBinaryTree.java[Java]
-//|{doc_base_url}/1104-path-in-zigzag-labelled-binary-tree.adoc[题解]
-//|Medium
-//|
-//
+
+|{counter:codes}
+|{leetcode_base_url}/path-in-zigzag-labelled-binary-tree/[1104. Path In Zigzag Labelled Binary Tree^]
+|{source_base_url}/_1104_PathInZigzagLabelledBinaryTree.java[Java]
+|{doc_base_url}/1104-path-in-zigzag-labelled-binary-tree.adoc[题解]
+|Medium
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/filling-bookcase-shelves/[1105. Filling Bookcase Shelves^]
 //|{source_base_url}/_1105_FillingBookcaseShelves.java[Java]

docs/1104-path-in-zigzag-labelled-binary-tree.adoc

@@ -1,45 +1,70 @@
 [#1104-path-in-zigzag-labelled-binary-tree]
-= 1104. Path In Zigzag Labelled Binary Tree
+= 1104. 二叉树寻路
 
-{leetcode}/problems/path-in-zigzag-labelled-binary-tree/[LeetCode - Path In Zigzag Labelled Binary Tree^]
+https://leetcode.cn/problems/path-in-zigzag-labelled-binary-tree/[LeetCode - 1104. 二叉树寻路 ^]
 
-In an infinite binary tree where every node has two children, the nodes are labelled in row order.
+在一棵无限的二叉树上，每个节点都有两个子节点，树中的节点 *逐行* 依次按 “之” 字形进行标记。
 
-In the odd numbered rows (ie., the first, third, fifth,...), the labelling is left to right, while in the even numbered rows (second, fourth, sixth,...), the labelling is right to left.
+如下图所示，在奇数行（即，第一行、第三行、第五行……）中，按从左到右的顺序进行标记；
 
-image::https://assets.leetcode.com/uploads/2019/06/24/tree.png[{image_attr}]
+而偶数行（即，第二行、第四行、第六行……）中，按从右到左的顺序进行标记。
 
-Given the `label` of a node in this tree, return the labels in the path from the root of the tree to the node with that `label`.
+image::images/1104-01.png[{image_attr}]
 
- 
-*Example 1:*
+给你树上某一个节点的标号 `label`，请你返回从根节点到该标号为 `label` 节点的路径，该路径是由途经的节点标号所组成的。
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* label = 14
-*Output:* [1,3,4,14]
-----
+*示例 1：*
 
-*Example 2:*
+....
+输入：label = 14
+输出：[1,3,4,14]
+....
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* label = 26
-*Output:* [1,2,6,10,26]
-----
+*示例 2：*
+
+....
+输入：label = 26
+输出：[1,2,6,10,26]
+....
 
- 
-*Constraints:*
 
+*提示：*
 
-* `1 <= label <= 10^6`
+* `1 \<= label \<= 10^6^`
 
 
 
+== 思路分析
+
+找到每一层的左右端点的值，确认所求数值在本层中的下标（从 `1` 开始），开始循环，每次循环下标都除以 `2` 再向上取整。这样就可以获取每个数值对应的节点。
+
+看官方题解，每层两端的数值都可以从当前层数获取，不需要单独存储。
 
 [[src-1104]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_1104_PathInZigzagLabelledBinaryTree.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_1104_PathInZigzagLabelledBinaryTree_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
 
+. https://leetcode.cn/problems/path-in-zigzag-labelled-binary-tree/solutions/901472/er-cha-shu-xun-lu-by-leetcode-solution-ryx0/[1104. 二叉树寻路 - 官方题解^]
+. https://leetcode.cn/problems/path-in-zigzag-labelled-binary-tree/solutions/902506/gong-shui-san-xie-yi-ti-shuang-jie-mo-ni-rw2d/[1104. 二叉树寻路 - 一题双解 :「模拟」&「数学」^]
+. https://leetcode.cn/problems/path-in-zigzag-labelled-binary-tree/solutions/96636/jian-dan-yi-dong-de-gong-shi-shi-jian-guo-100-by-a/[1104. 二叉树寻路 - 简单易懂的公式 时间过100%^]

docs/images/1104-01.png

@@ -2291,7 +2291,7 @@ include::1094-car-pooling.adoc[leveloffset=+1]
 
 // include::1103-distribute-candies-to-people.adoc[leveloffset=+1]
 
-// include::1104-path-in-zigzag-labelled-binary-tree.adoc[leveloffset=+1]
+include::1104-path-in-zigzag-labelled-binary-tree.adoc[leveloffset=+1]
 
 // include::1105-filling-bookcase-shelves.adoc[leveloffset=+1]
 

docs/index.adoc

@@ -903,6 +903,11 @@ endif::[]
 |{doc_base_url}/0986-interval-list-intersections.adoc[题解]
 |✅ 双指针。先排除两种没有交集的情况，剩下四种有交集的情况，左边选两个里面最大的，右边选两个里面最小的。哪个小，向右移动哪个的指针。
 
+|{counter:codes2503}
+|{leetcode_base_url}/path-in-zigzag-labelled-binary-tree/[1104. 二叉树寻路^]
+|{doc_base_url}/1104-path-in-zigzag-labelled-binary-tree.adoc[题解]
+|✅ 找到每一层的左右端点的值，确认所求数值在本层中的下标（从 `1` 开始），开始循环，每次循环下标都除以 `2` 再向上取整。这样就可以获取每个数值对应的节点。
+
 
 |===
 

logbook/202503.adoc

@@ -0,0 +1,57 @@
+package com.diguage.algo.leetcode;
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class _1104_PathInZigzagLabelledBinaryTree {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-06-03 15:52:57
+   */
+  public List<Integer> pathInZigZagTree(int label) {
+    int cur = label;
+    int len = 1;
+    boolean isOdd;
+    int max = 0;
+    int depth = ((int) (Math.log(label) / Math.log(2))) + 1;
+    int[] left = new int[depth];
+    int[] right = new int[depth];
+    int level = 1;
+    while (left[depth - 1] == 0) {
+      isOdd = (level & 1) == 1;
+      max += len;
+      if (isOdd) {
+        right[level - 1] = max;
+        left[level - 1] = max - len + 1;
+      } else {
+        left[level - 1] = max;
+        right[level - 1] = max - len + 1;
+      }
+      if (cur > len) {
+        cur -= len;
+      }
+      level++;
+      len *= 2;
+    }
+    List<Integer> result = new ArrayList<>();
+    int index = depth - 1;
+    boolean isEven = (depth & 1) == 0;
+    int levelIndex = isEven ? (left[depth - 1] - right[depth - 1] + 1) - cur + 1 : cur;
+    while (index >= 0) {
+      if (left[index] < right[index]) {
+        result.add(left[index] + levelIndex - 1);
+      } else {
+        result.add(left[index] - levelIndex + 1);
+      }
+      levelIndex = (int) Math.ceil(levelIndex / 2.0);
+      index--;
+    }
+    return result.reversed();
+  }
+  // end::answer[]
+  public static void main(String[] args) {
+    new _1104_PathInZigzagLabelledBinaryTree()
+      .pathInZigZagTree(26);
+  }
+}