105. Construct Binary Tree from Preorder and Inorder Traversal

难度: Medium

题目描述

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.

 

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.

解题思路

代码实现

解决方案

java

/**
 * 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 {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        Map<Integer,Integer> inMap  = new HashMap<>();
        for(int i=0;i<inorder.length;i++){
            inMap.put(inorder[i],i);
        }

        return builder(preorder,inorder,0,preorder.length-1,0,inorder.length-1,inMap);

    }

    public TreeNode builder(int[] preorder, int[] inorder, int preLeft, int preRight, int inLeft, int inRight,
            Map<Integer, Integer> inMap) {

        if (preLeft > preRight) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[preLeft]);

        int rootInOrderIndex = inMap.get(preorder[preLeft]);
        int leftLength = rootInOrderIndex - inLeft;
        int rightLength = inRight - rootInOrderIndex;

        TreeNode left = builder(preorder, inorder, preLeft + 1, preLeft + leftLength, inLeft, rootInOrderIndex - 1,
                inMap);
        TreeNode right = builder(preorder, inorder, preLeft + leftLength + 1, preRight, rootInOrderIndex + 1, inRight,
                inMap);
        root.left = left;
        root.right = right;
        return root;

    }
}