/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
int n = postorder.length;
if(n == 0){
return null;
}
int leftSize = indexOf(inorder, postorder[n-1]);// 左区间的长度
int[] in1 = Arrays.copyOfRange(inorder, 0, leftSize);
int[] in2 = Arrays.copyOfRange(inorder, leftSize+1, n);// 记得排除中间的根节点
int[] post1 = Arrays.copyOfRange(postorder, 0, leftSize);
int[] post2 = Arrays.copyOfRange(postorder, leftSize, n-1);// 排除最后一个根节点
TreeNode left = buildTree(in1, post1);
TreeNode right = buildTree(in2,post2);
return new TreeNode(postorder[n-1],left,right);
}
// 返回 x 在 a 中的下标,保证 x 一定在 a 中
private int indexOf(int[] a , int x){
for(int i = 0; ; i++){
if(a[i] == x){
return i;
}
}
}
}