Construct Binary Tree from Inorder and Postorder Traversal

標籤：

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

 Notice

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

 Example

Given inorder [1,2,3] and postorder [1,3,2], return a tree:

  2 / 1   3

分析：這是非常典型的遞迴問題，postorder的最後一個是root，在inorder裡面找出root的位置，左邊部分為左子樹，右邊部分為右子樹，稍微麻煩的部分就是確定左子樹和右子樹的starting index。

 1 /** 2  * Definition of TreeNode: 3  * public class TreeNode { 4  *     public int val; 5  *     public TreeNode left, right; 6  *     public TreeNode(int val) { 7  *         this.val = val; 8  *         this.left = this.right = null; 9  *     }10  * }11  */12  13  14 public class Solution {15     /**16      *@param inorder : A list of integers that inorder traversal of a tree17      *@param postorder : A list of integers that postorder traversal of a tree18      *@return : Root of a tree19      */20     public TreeNode buildTree(int[] inorder, int[] postorder) {21         if (inorder == null || postorder == null || postorder.length == 0 || inorder.length != postorder.length) return null;22         23         return buildTree(inorder, 0, postorder, 0, inorder.length);24         25     }26     27     public TreeNode buildTree(int[] inorder, int inStart, int[] postorder, int postStart, int length) {28         if (length <= 0) return null;29         TreeNode root = new TreeNode(postorder[postStart + length - 1]);30 31         int index = findIndex(inorder, root.val);32         root.left = buildTree(inorder, inStart, postorder, postStart, index - inStart);33         root.right = buildTree(inorder, index + 1, postorder, postStart + (index - inStart), length - (index - inStart) - 1);34 35         return root;    36         37     }38 39     public int findIndex(int[] inorder, int val) {40         for (int i = 0; i < inorder.length; i++) {41             if (inorder[i] == val) {42                 return i;43             }44         }45         return -1;46     }47 }

 

  

Construct Binary Tree from Inorder and Postorder Traversal