Construct Binary Tree from Preorder and Inorder Traversal

Construct Binary Tree from Preorder and Inorder Traversal

问题:

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

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

思路:

  dfs

我的代码:

public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(inorder == null || inorder.length == 0)  return null;
        if(inorder.length == 1) return new TreeNode(inorder[0]);
        int len = preorder.length;
        int target = preorder[0];
        int index = getIndex(inorder, target);
        TreeNode root = new TreeNode(target);
        root.left = buildTree(Arrays.copyOfRange(preorder,1,index + 1),Arrays.copyOfRange(inorder,0,index));
        root.right = buildTree(Arrays.copyOfRange(preorder, index + 1, len),Arrays.copyOfRange(inorder, index+1, len));
        return root;
    }
    public int getIndex(int[] inorder, int target)
    {
        for(int i = 0; i < inorder.length; i++)
        {
            if(inorder[i] == target)
                return i;
        }
        return -1;
    }
}
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原文地址:https://www.cnblogs.com/sunshisonghit/p/4337358.html