Medium
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
思路跟之间拿到inorder + preorder construct binary tree基本是一样的思路。根据postorder里最后面的是root的规律，我们维持一个postIndex从后面往前依次遍历postorder. 每次指向某个postorder里的元素，那我们就构建一个以该元素的值为val的tNode. 再去inorder里面找到这个tNode.val， 得到inIndex. 这样我们就可以把inorder里剩下的元素分为tNode的左右子树。inorder里在inIndex前面的在tNode的左子树，inIndex后面的在tNode的右子树。那么我们就可以分别在左右两边依次递归，要注意每次递归锁定的范围，可以用start, end来固定，并注意每次更新的规律。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
int postIndex = 0;
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (inorder == null || postorder == null || inorder.length == 0 || postorder.length == 0){
return null;
}
postIndex = postorder.length - 1;
return helper(inorder, postorder, 0, inorder.length - 1);
}
private TreeNode helper(int[] inorder, int[] postorder, int start, int end){
if (start > end || postIndex < 0){
return null;
}
TreeNode tNode = new TreeNode(postorder[postIndex--]);
int inIndex = search(inorder, tNode.val, start, end);
tNode.right = helper(inorder, postorder, inIndex + 1, end);
tNode.left = helper(inorder, postorder, start, inIndex - 1);
return tNode;
}
private int search(int[] inorder, int target, int left, int right){
for (int i = left; i <= right; i++){
if (inorder[i] == target){
return i;
}
}
return -1;
}
}
