TraversalTest__从中序与后序遍历序列构造二叉树

Author: CPU-Code

leetcode/README.md

@@ -67,6 +67,7 @@
 - [x] [SumTest__路径总和](src/main/java/com/cpucode/binary/tree/path/sum/SumTest.java)
 - [x] [SumTest2__路径总和2](src/main/java/com/cpucode/binary/tree/path/sum/SumTest2.java)
 - [ ] [TraversalTest__从前序与中序遍历序列构造二叉树](src/main/java/com/cpucode/binary/tree/preorder/inorder/TraversalTest.java)
+- [ ] [TraversalTest__从中序与后序遍历序列构造二叉树](src/main/java/com/cpucode/binary/tree/inorder/postorder/TraversalTest.java)
 
 ------------------------
 

leetcode/src/main/java/com/cpucode/binary/tree/inorder/postorder/TraversalTest.java

@@ -0,0 +1,79 @@
+package com.cpucode.binary.tree.inorder.postorder;
+
+import java.util.HashMap;
+import java.util.Map;
+
+/**
+ * 106. 从中序与后序遍历序列构造二叉树
+ *
+ * 题目描述
+ * 根据一棵树的中序遍历与后序遍历构造二叉树。
+ *
+ * 注意:
+ * 你可以假设树中没有重复的元素。
+ *
+ * 例如，给出
+ * 中序遍历 inorder = [9,3,15,20,7]
+ * 后序遍历 postorder = [9,15,7,20,3]
+ * 返回如下的二叉树：
+ *
+ *     3
+ *    / \
+ *   9  20
+ *     /  \
+ *    15   7
+ * 解法
+ * 思路同 105。
+ *
+ * @author : cpucode
+ * @date : 2021/5/18
+ * @time : 22:50
+ * @github : https://github.com/CPU-Code
+ * @csdn : https://blog.csdn.net/qq_44226094
+ */
+public class TraversalTest {
+    private Map<Integer, Integer> indexes = new HashMap<>();
+
+    public static void main(String[] args) {
+
+    }
+
+    public TreeNode buildTree(int[] inorder, int[] postorder) {
+        int n = inorder.length;
+
+        for (int i = 0; i < n; ++i) {
+            indexes.put(inorder[i], i);
+        }
+
+        return build(inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);
+    }
+
+    private TreeNode build(int[] inorder, int[] postorder, int i1, int i2, int p1, int p2) {
+        if (i1 > i2 || p1 > p2) {
+            return null;
+        }
+        int rootVal = postorder[p2];
+        int pos = indexes.get(rootVal);
+        TreeNode root = new TreeNode(rootVal);
+
+        root.left = pos == i1 ? null : build(inorder, postorder, i1, pos - 1, p1, p1 - i1 + pos - 1);
+        root.right = pos == i2 ? null : build(inorder, postorder, pos + 1, i2, p1 - i1 + pos, p2 - 1);
+
+        return root;
+    }
+}
+
+class TreeNode {
+    int val;
+    TreeNode left;
+    TreeNode right;
+
+    TreeNode() {}
+    TreeNode(int val) { this.val = val; }
+    TreeNode(int val, TreeNode left, TreeNode right) {
+        this.val = val;
+        this.left = left;
+        this.right = right;
+    }
+}
+