Problem 105

Construct Binary Tree from Preorder and Inorder Traversal

Posted by Ruizhi Ma on August 23, 2019

问题描述

解决思路

问题解惑

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        //corner case
        if(preorder == null || inorder == null || preorder.length == 0 || preorder.length != inorder.length) return null;
        
        //pre_end is unnecessary since in_end is enough
        return helper(preorder, inorder, 0, 0, inorder.length - 1);
    }
    
    public TreeNode helper(int[] preorder, int[] inorder, int pre_st, int in_st, int in_end){
        if(pre_st > preorder.length || in_st > in_end) return null;
        //the first node in preorder is the root
        TreeNode current = new TreeNode(preorder[pre_st]);
        //to find the index of root node in inorder
        int i = in_st;
        while(i <= in_end){
            if(inorder[i] == preorder[pre_st]) break;
            i++;
        }
        
        current.left = helper(preorder, inorder, pre_st + 1, in_st, i - 1);
        current.right = helper(preorder, inorder, pre_st + (i - in_st + 1), i + 1, in_end);
        
        return current;
    }
}

复杂度分析

  1. 时间复杂度:2^n
  2. 空间复杂度:O(1)