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

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:
    3
   / 
  9  20
    /  
   15   7

从中序和后序遍历序列构造二叉树。

题目就是题意，影子题105，几乎一模一样。

思路还是递归/分治。因为有了后序遍历所以后序遍历的最后一个 node 就是根节点。有了根节点，可以依据这个根节点在 inorder 遍历的位置，将 inorder 的结果分成左子树和右子树。代码中的 index 变量，找的是根节点在 inorder 里面的坐标。照着例子解释一下，

inorder = [9,3,15,20,7] - 9是左子树，15，20，7是右子树
postorder = [9,15,7,20,3] - 3是根节点

思路和代码可以参照105题，几乎是一样的。

时间O(n)

空间O(n)

Java实现

 1 /**
 2 * Definition for a binary tree node.
 3 * public class TreeNode {
 4 *     int val;
 5 *     TreeNode left;
 6 *     TreeNode right;
 7 *     TreeNode(int x) { val = x; }
 8 * }
 9 */
10 class Solution {
11     public TreeNode buildTree(int[] inorder, int[] postorder) {
12         return helper(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
13     }
14 
15     private TreeNode helper(int[] inorder, int inStart, int inEnd, int[] postorder, int pStart, int pEnd) {
16         // corner case
17         if (inStart > inEnd || pStart > pEnd) {
18             return null;
19         }
20         TreeNode root = new TreeNode(postorder[pEnd]);
21         // 找根节点在inorder的位置
22         int index = 0;
23         while (inorder[inStart + index] != postorder[pEnd]) {
24             index++;
25         }
26         root.left = helper(inorder, inStart, inStart + index - 1, postorder, pStart, pStart + index - 1);
27         root.right = helper(inorder, inStart + index + 1, inEnd, postorder, pStart + index, pEnd - 1);
28         return root;
29     }
30 }