LeetCode[106]从中序和后序遍历序列构造二叉树

思路：

我觉得这种题还是要找好边界，这道题和从中序和前序遍历序列构造二叉树差不多，就是后序遍历和前序遍历是反着来的，后序遍历最后一个是头节点，然后递归时中序遍历的处理逻辑没什么变化，唯一有变化的是后序遍历的递归逻辑，在后序遍历中确认左子树和右子树的范围，左子树范围是头节点---头节点+左子树长度-1，右子树范围头节点+左子树长度---尾节点-1。

代码：

/*** 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 {Map<Integer, Integer> map;public TreeNode buildTree(int[] inorder, int[] postorder) {map = new HashMap<>();for (int i = 0; i < inorder.length; i++) {map.put(inorder[i], i);}return helper(inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);}public TreeNode helper(int[] inorder, int[] postorder, int i_start, int i_end, int p_start, int p_end) {if (p_start > p_end)return null;TreeNode root = new TreeNode(postorder[p_end]);int mid = map.get(postorder[p_end]);int leftLength = mid - i_start;root.left = helper(inorder, postorder, i_start, mid - 1, p_start, p_start + leftLength - 1);root.right = helper(inorder, postorder, mid + 1, i_end, p_start + leftLength, p_end - 1);return root;}
}