[add] 106. Construct Binary Tree from Inorder and Postorder Traversal

Author: brianway

algorithms-leetcode/README.md

@@ -62,6 +62,7 @@ Leetcode problems classified by company:
 |102|Binary Tree Level Order Traversal|Medium|Binary Tree,DFS/BFS|一题多解| 
 |104|Maximum Depth of Binary Tree|Easy|Binary Tree/Queue||
 |105|Construct Binary Tree from Preorder and Inorder Traversal|Medium|Binary Search,divide and conquer|只写了常规解法|
+|106|Construct Binary Tree from Inorder and Postorder Traversal|Medium|Binary Search,divide and conquer|只写了常规解法|
 |107|Binary Tree Level Order Traversal II|Medium|Binary Tree,DFS/BFS|同第102题,TODO| 
 |110|Balanced Binary Tree|Easy|Binary Tree/Queue||
 |111|Minimum Depth of Binary Tree|Easy|Binary Tree/Queue||

algorithms-leetcode/src/main/java/com/brianway/learning/algorithms/leetcode/medium/ConstructBinaryTreeFromInorderAndPostorderTraversal.java

@@ -0,0 +1,78 @@
+package com.brianway.learning.algorithms.leetcode.medium;
+
+import com.brianway.learning.algorithms.leetcode.common.TreeNode;
+
+/**
+ * 106. Construct Binary Tree from Inorder and Postorder Traversal
+ * Question: https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
+ * 关键题设：inorder and postorder consist of unique values.
+ *
+ * @auther brian
+ * @since 2022/9/7 21:08
+ */
+public class ConstructBinaryTreeFromInorderAndPostorderTraversal {
+    public TreeNode buildTree(int[] inorder, int[] postorder) {
+        return null;
+    }
+
+    /**
+     * 递归
+     */
+    public class ConstructBinaryTreeFromInorderAndPostorderTraversal0 extends ConstructBinaryTreeFromInorderAndPostorderTraversal {
+        @Override
+        public TreeNode buildTree(int[] inorder, int[] postorder) {
+            return buildTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
+        }
+
+        /**
+         * root=postorder[postHigh], 在inorder中找到inRootIndex, 则
+         * inorder左子树：[inLow, inRootIndex-1],
+         * inorder右子树： [inRootIndex+1, inHigh]
+         * <p>
+         * 下面确定postorder的左右子树的数组下标边界
+         * 左子树的size=inRootIndex-inLow
+         * 右子树的size=inHigh-(inRootIndex+1)+1=inHigh-inRootIndex
+         * <p>
+         * postorder左子树：[postLow, postLow+左子树的size-1]
+         * postorder右子树：[postHigh-1-右子树的size+1 ,postHigh-1]
+         * <p>
+         * 暂留：
+         * 又（postLow+左子树的size-1） +1= postHigh-1-右子树的size+1
+         * postLow+左子树的size= postHigh-右子树的size
+         * postLow+inRootIndex-inLow=postHigh-（inHigh-inRootIndex）
+         * postLow-inLow=postHigh-inHigh
+         * <p>
+         * <p>
+         * 时间复杂度 O(n log n)  假设二叉树层数为k, 每一层找inorder的根节点下标都平均需要遍历 n/2,  所以 时间复杂度 O(k*n),
+         * 极端情况下，k=n，且每一层遍历都是n, 时间复杂度退化为  O(n^2)
+         * 空间复杂度 O(1)
+         */
+        public TreeNode buildTree(int[] inorder, int inLow, int inHigh, int[] postorder, int postLow, int postHigh) {
+            // 终止条件
+            if (inHigh < inLow) {
+                return null;
+            }
+            if (inHigh == inLow) {
+                return new TreeNode(inorder[inHigh]);
+            }
+
+            int root = postorder[postHigh];
+            // find inRootIndex
+            int inRootIndex = findInorderRootIndex(root, inorder, inLow, inHigh);
+            TreeNode left = buildTree(inorder, inLow, inRootIndex - 1,
+                    postorder, postLow, postLow + inRootIndex - inLow - 1);
+            TreeNode right = buildTree(inorder, inRootIndex + 1, inHigh,
+                    postorder, postHigh - (inHigh - inRootIndex), postHigh - 1);
+            return new TreeNode(root, left, right);
+        }
+    }
+
+    public int findInorderRootIndex(int root, int[] inorder, int low, int high) {
+        for (int i = low; i <= high; i++) {
+            if (inorder[i] == root) {
+                return i;
+            }
+        }
+        throw new IllegalArgumentException("not found");
+    }
+}