Construct Binary Tree from Preorder and Inorder Traversal

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分类专栏： leetcode java 文章标签： Java leetcode

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给定一棵树的前序和中序遍历，使用递归方法来构造这棵二叉树，注意树中没有重复元素。

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

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

本题采用递归的方式来进行：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder == null || inorder == null){
            return null;
        }
        HashMap<Integer,Integer> map = new HashMap<>();
        for(int i = 0; i < inorder.length;i++){
            map.put(inorder[i],i);
        }
        return  recursion(preorder,inorder,0, preorder.length-1,0,inorder.length-1,map);
    }
     public TreeNode recursion(int[] preorder, int[] inorder, int preLeft, int preRight, int inLeft,  int inRight, HashMap<Integer, Integer> map){
        if(preLeft > preRight || inLeft > inRight){
            return null;
        }
        TreeNode root = new TreeNode(preorder[preLeft]);
        int index = map.get(root.val);
        root.left = recursion(preorder,inorder,preLeft+1, index-inLeft+preLeft,inLeft,index-1,map);
        root.right = recursion(preorder,inorder,index-inLeft+preLeft+1,preRight,index+1,inRight,map);
        return root;
    }
}