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| 1 | +/* |
| 2 | + * @lc app=leetcode.cn id=106 lang=java |
| 3 | + * |
| 4 | + * [106] 从中序与后序遍历序列构造二叉树 |
| 5 | + * |
| 6 | + * https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/ |
| 7 | + * |
| 8 | + * algorithms |
| 9 | + * Medium (68.64%) |
| 10 | + * Likes: 230 |
| 11 | + * Dislikes: 0 |
| 12 | + * Total Accepted: 39.9K |
| 13 | + * Total Submissions: 58K |
| 14 | + * Testcase Example: '[9,3,15,20,7]\n[9,15,7,20,3]' |
| 15 | + * |
| 16 | + * 根据一棵树的中序遍历与后序遍历构造二叉树。 |
| 17 | + * |
| 18 | + * 注意: |
| 19 | + * 你可以假设树中没有重复的元素。 |
| 20 | + * |
| 21 | + * 例如,给出 |
| 22 | + * |
| 23 | + * 中序遍历 inorder = [9,3,15,20,7] |
| 24 | + * 后序遍历 postorder = [9,15,7,20,3] |
| 25 | + * |
| 26 | + * 返回如下的二叉树: |
| 27 | + * |
| 28 | + * 3 |
| 29 | + * / \ |
| 30 | + * 9 20 |
| 31 | + * / \ |
| 32 | + * 15 7 |
| 33 | + * |
| 34 | + * |
| 35 | + */ |
| 36 | + |
| 37 | +// @lc code=start |
| 38 | +/** |
| 39 | + * Definition for a binary tree node. |
| 40 | + * public class TreeNode { |
| 41 | + * int val; |
| 42 | + * TreeNode left; |
| 43 | + * TreeNode right; |
| 44 | + * TreeNode(int x) { val = x; } |
| 45 | + * } |
| 46 | + */ |
| 47 | +class Solution { |
| 48 | + private Map<Integer, Integer> map = new HashMap<>(); |
| 49 | + |
| 50 | + public TreeNode buildTree(int[] inorder, int[] postorder) { |
| 51 | + if (inorder == null || postorder == null || inorder.length != postorder.length) { |
| 52 | + return null; |
| 53 | + } |
| 54 | + |
| 55 | + for (int i = 0; i < inorder.length; i++) { |
| 56 | + map.put(inorder[i], i); |
| 57 | + } |
| 58 | + |
| 59 | + return helper(postorder, postorder.length - 1, 0, inorder.length - 1); |
| 60 | + } |
| 61 | + |
| 62 | + private TreeNode helper(int[] postorder, int postIndex, int inStart, int inEnd) { |
| 63 | + if (postIndex < 0 || inStart > inEnd) { |
| 64 | + return null; |
| 65 | + } |
| 66 | + |
| 67 | + TreeNode root = new TreeNode(postorder[postIndex]); |
| 68 | + int inIndex = map.get(postorder[postIndex]); |
| 69 | + int rightTreeSize = inEnd - inIndex; |
| 70 | + |
| 71 | + root.left = helper(postorder, postIndex - rightTreeSize - 1, inStart, inIndex - 1); |
| 72 | + root.right = helper(postorder, postIndex - 1, inIndex + 1, inEnd); |
| 73 | + |
| 74 | + return root; |
| 75 | + } |
| 76 | +} |
| 77 | +// @lc code=end |
| 78 | + |
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