LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal

原题链接在这里：https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

题目：

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

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

题解：

这道题与Construct Binary Tree from Preorder and Inorder Traversal思路相似，不同之处在于这里的root在postorder的最有一位，其他都相同，由inorder找出分切点.

Time Complexity: O(n). Space: O(n).

AC Java:

/**
 * 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) {
        if(inorder == null || inorder.length == 0 || postorder == null || postorder.length == 0){
            return null;
        }
        HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
        for(int i = 0; i<inorder.length; i++){
            hm.put(inorder[i], i);
        }
        return helper(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1, hm);
    }
    
    private TreeNode helper(int [] inorder, int inL, int inR, int [] postorder, int posL, int posR, HashMap<Integer, Integer> hm){
        if(inL > inR || posL > posR){
            return null;
        }
        TreeNode root = new TreeNode(postorder[posR]);
        int rootIndex = hm.get(postorder[posR]);
        root.left = helper(inorder, inL, rootIndex-1, postorder, posL, posL+rootIndex-inL-1, hm);
        root.right = helper(inorder, rootIndex+1, inR, postorder, posL+rootIndex-inL, posR-1, hm);
        return root;
    }
}

类似Construct Binary Tree from Preorder and Postorder Traversal.