leetcode_106_Construct Binary Tree from Inorder and Postorder Traversal

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描述：

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

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

思路：

1.将中序遍历序列和其对应的下标存储到一个map中，方便下面的查找

2.递归选取后序序列的倒数第一个元素作为树的根节点，然后查找根节点在后序序列中位置inorderIndex，endInorder-inorderIndex可以得到右子树的长度

3.根据右子树的长度和endPreOrder可以求出后序序列中右子树的起始位置

4.从上面可以求出左右子树的后序序列和中序序列的起始位置，递归调用建树过程即可。

代码：

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    int[] ArrPostorder;int[] ArrInorder;Map<Integer, Integer> mapInorder = new HashMap<Integer, Integer>();public TreeNode buildTree(int[] inorder, int[] postorder) {if (inorder.length == 0 || inorder == null)return null;ArrPostorder = postorder;ArrInorder = inorder;for (int i = 0; i < inorder.length; i++)mapInorder.put(inorder[i], i);int start = 0, end = postorder.length - 1;TreeNode root = new TreeNode(0);createTree(root, start, end, start, end);return root;}public void createTree(TreeNode root, int start1, int end1, int start2,int end2) {int subStart1,subStart2,subEnd1,subEnd2;int target = ArrPostorder[end2];int indexInOrder = mapInorder.get(target);int len = end1-indexInOrder;int indexPostOrder = end2-len;if (start1 <= indexInOrder - 1) {subEnd1 = indexInOrder-1;subEnd2 = indexPostOrder - 1;root.left = new TreeNode(0);createTree(root.left, start1, subEnd1, start2, subEnd2);}root.val = target;if (indexInOrder + 1 <= end1) {subStart1 = indexInOrder + 1;subStart2 = indexPostOrder;subEnd2=end2-1;root.right = new TreeNode(0);createTree(root.right, subStart1, end1, subStart2, subEnd2);}return;}}

结果：

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