106.Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal

106. Construct Binary Tree from Inorder and Postorder Traversal

题目地址

https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

题目描述

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

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

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

代码

Approach #1 Recursion

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
  int post_idx;
  int[] postorder;
  HashMap<Integer, Integer> idx_map = new HashMap();
  public TreeNode buildTree(int[] inorder, int[] postorder) {
        this.postorder = postorder;
    post_idx = postorder.length - 1;

    int idx = 0;
    for (Integer val: inorder) {
      idx_map.put(val, idx++);
    }
    return helper(0, inorder.length - 1);
  }

  public TreeNode helper(int in_left, int in_right) {
    if (in_left > in_right) { // inorder的作用
      return null;
    }

    int root_val = postorder[post_idx];
    TreeNode root = new TreeNode(root_val);

    int index = idx_map.get(root_val); // inorder的作用

    post_idx--;
    root.left = helper(in_left, index - 1);
    root.right = helper(index + 1, in_right);

    return root;
  }
}