java_build_binary_tree_from_preOrder_inOrder

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

Examples

Example 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

Example 2:

Input: preorder = [-1], inorder = [-1]
Output: [-1]

Constraints:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• 3000 <= preorder[i], inorder[i] <= 3000
• preorder and inorder consist of unique values.
• Each value of inorder also appears in preorder.
• preorder is guaranteed to be the preorder traversal of the tree.
• inorder is guaranteed to be the inorder traversal of the tree.

解析

題目以 Pre-Order Traversal 與 In-Order Traversal 給出兩個整數陣列

求實作一個演算法建立出這棵二元樹

Pre-Order Traversal 特性會有以下走訪順序如下：

二元樹root , root.Left 子樹, root.Right 子樹

In-Order Traversal 特性會有以下走訪順序如下：

root.Left 子樹, 二元樹root, root.Right 子樹

觀察下圖

可以發現

每個 pre-order的點先遇到的都是子樹的跟結點

透過 pre-order的點去找到在 in-order的位置剛好可以把左右子數分開

所以可以透過遞迴關係來建立二元樹

程式碼

class Solution {
  /**
   * 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;
   *     }
   * }
   */
  int preorderIdx;
  public TreeNode buildTree(int[] preorder, int[] inorder) {
    preorderIdx = 0;
    HashMap<Integer, Integer> inorderIdxMap = new HashMap<>();
    for (int idx = 0; idx < inorder.length; idx++) {
      inorderIdxMap.put(inorder[idx], idx);
    }
    return arrayToTree(preorder, 0, preorder.length - 1, inorderIdxMap);
  }
  public TreeNode arrayToTree(int[] preorder, int left, int right, HashMap<Integer, Integer> inorderIdxMap) {
    if (left > right) {
      return null;
    }
    int rootValue = preorder[preorderIdx];
    preorderIdx++;
    TreeNode root = new TreeNode(rootValue);
    root.left = arrayToTree(preorder, left, inorderIdxMap.get(rootValue) - 1, inorderIdxMap);
    root.right = arrayToTree(preorder,inorderIdxMap.get(rootValue) + 1, right, inorderIdxMap);
    return root;
  }
}

困難點

1. 看出 Pre-Order Traversal 與 In-Order Traversal 的特性
2. 看出遞迴關係

Solve Point

• Understand what problem need to solve
• Analysis Complexity