Construct Binary Tree from Inorder and Postorder Traversal

题意：

给定一棵树的中序和后续遍历序列，还原出这棵树；

分析：

简单，后续遍历的最后一个元素肯定是根节点，然后再中序遍历中找到根节点，此节点之前的元素必属于根节点左子树，之后的元素必定属于根节点右子树；然后再后序遍历序列中找到相同长度的左子树和右子树的序列，递归。。。

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
int n = inorder.length;

return buildTree(inorder, postorder, 0, n, 0, n);

}
public TreeNode buildTree(int[] inorder, int[] postorder, int i1, int i2, int p1, int p2) {
if(i1 >= i2){
return null;
}

int value = postorder[p2 - 1];
TreeNode root = new TreeNode(value);

int i;
for(i = i1;i < i2; i++){
if(inorder[i] == value){
break;
}
}

int length = i - i1;

root.left = buildTree(inorder, postorder, i1, i, p1, p1 + length);
root.right = buildTree(inorder, postorder, i+1, i2, p1 + length, p2 - 1);

return root;
}

}