再次完成 111

Author: diguage

docs/0111-minimum-depth-of-binary-tree.adoc

@@ -1,3 +1,4 @@
+[#0111-minimum-depth-of-binary-tree]
 = 111. Minimum Depth of Binary Tree
 
 https://leetcode.com/problems/minimum-depth-of-binary-tree/[LeetCode - Minimum Depth of Binary Tree]
@@ -42,3 +43,18 @@ return its minimum depth = 2.
 include::{sourcedir}/_0111_MinimumDepthOfBinaryTree.java[]
 ----
 
+
+[{java_src_attr}]
+----
+include::{sourcedir}/_0111_MinimumDepthOfBinaryTree_2.java[]
+----
+
+
+TIP: 可以换个角度理解这道题：使用 Morris 遍历树时，遍历到每个节点时，计算该节点的深度，然后从中筛选出叶子阶段的深度，取最小值即可。
+
+[{java_src_attr}]
+----
+include::{sourcedir}/_0111_MinimumDepthOfBinaryTree_Morris.java[]
+----
+
+竟然在 LeetCode 中文站点没有找到使用 Morris 遍历方法处理的解答。

src/main/java/com/diguage/algorithm/leetcode/_0111_MinimumDepthOfBinaryTree_2.java

@@ -0,0 +1,42 @@
+package com.diguage.algorithm.leetcode;
+
+import com.diguage.algorithm.util.TreeNode;
+
+import java.util.Objects;
+
+import static com.diguage.algorithm.util.TreeNodeUtils.buildTree;
+import static java.util.Arrays.asList;
+
+/**
+ * = 111. Minimum Depth of Binary Tree
+ *
+ * https://leetcode.com/problems/minimum-depth-of-binary-tree/[Minimum Depth of Binary Tree - LeetCode]
+ *
+ * @author D瓜哥, https://www.diguage.com/
+ * @since 2020-02-07 21:26
+ */
+public class _0111_MinimumDepthOfBinaryTree_2 {
+
+    public int minDepth(TreeNode root) {
+      if (Objects.isNull(root)) {
+        return 0;
+      }
+      if (root.left == null && root.right == null) {
+        return 1;
+      }
+      int result = Integer.MAX_VALUE;
+      if (Objects.nonNull(root.left)) {
+        result = Math.min(result, minDepth(root.left));
+      }
+      if (Objects.nonNull(root.right)) {
+        result = Math.min(result, minDepth(root.right));
+      }
+      return result + 1;
+    }
+
+    public static void main(String[] args) {
+        _0111_MinimumDepthOfBinaryTree_2 solution = new _0111_MinimumDepthOfBinaryTree_2();
+        int r1 = solution.minDepth(buildTree(asList(3, 9, 20, null, null, 15, 7)));
+        System.out.println((r1 == 2) + " : " + r1);
+    }
+}

src/main/java/com/diguage/algorithm/leetcode/_0111_MinimumDepthOfBinaryTree_Morris.java

@@ -0,0 +1,78 @@
+package com.diguage.algorithm.leetcode;
+
+import com.diguage.algorithm.util.TreeNode;
+
+import static com.diguage.algorithm.util.TreeNodeUtils.buildTree;
+import static java.util.Arrays.asList;
+
+/**
+ * = 111. Minimum Depth of Binary Tree
+ *
+ * https://leetcode.com/problems/minimum-depth-of-binary-tree/[Minimum Depth of Binary Tree - LeetCode]
+ *
+ * @author D瓜哥, https://www.diguage.com/
+ * @since 2020-02-07 21:26
+ */
+public class _0111_MinimumDepthOfBinaryTree_Morris {
+
+  /**
+   * 参考左程云《程序员代码面试指南》的解法
+   */
+  public int minDepth(TreeNode head) {
+    if (head == null) {
+      return 0;
+    }
+    TreeNode cur = head;
+    TreeNode mostRight = null;
+    int curLevel = 0;
+    int minHeight = Integer.MAX_VALUE;
+    while (cur != null) {
+      mostRight = cur.left;
+      if (mostRight != null) { // 当前 cur 有左子树，能到达两次
+        // cur 左子树上，右边界的节点个数
+        int leftTreeRightSize = 1;
+        // 找到 cur 左子树上最右边的节点
+        while (mostRight.right != null && mostRight.right != cur) {
+          leftTreeRightSize++;
+          mostRight = mostRight.right;
+        }
+        if (mostRight.right == null) {
+          // 第一次到达 cur，那么下一个节点的 level 必然+1
+          curLevel++;
+          mostRight.right = cur;
+          cur = cur.left;
+          continue;
+        } else {
+          // 第二次到达cur，那么下一个节点的 level = curLevel - leftTreeRightSize
+          // 此时检查 mostRight 是不是叶节点，记录答案
+          if (mostRight.left == null) {
+            minHeight = Math.min(minHeight, curLevel);
+          }
+          curLevel -= leftTreeRightSize;
+          mostRight.right = null;
+        }
+      } else {
+        // 当前 cur 没有左子树，只能到达有ICI，那么下一个节点的 level 必然+1
+        curLevel++;
+      }
+      cur = cur.right;
+    }
+    int finalRight = 1;
+    cur = head;
+    while (cur.right != null) {
+      finalRight++;
+      cur = cur.right;
+    }
+    // 最后不要忘了单独看一看整棵树的最右节点是不是叶节点
+    if (cur.left == null && cur.right == null) {
+      minHeight = Math.min(minHeight, finalRight);
+    }
+    return minHeight;
+  }
+
+  public static void main(String[] args) {
+    _0111_MinimumDepthOfBinaryTree_Morris solution = new _0111_MinimumDepthOfBinaryTree_Morris();
+    int r1 = solution.minDepth(buildTree(asList(3, 9, 20, null, null, 15, 7)));
+    System.out.println((r1 == 2) + " : " + r1);
+  }
+}