LeetCode—105. Construct Binary Tree from Preorder and Inorder Traversal

题目

根据前序遍历和中序遍历建立二叉树。注意二叉树中没有重复的数字。

思路及解法

根据前序遍历和中序遍历的特点，通过前序遍历列表确定根节点，在中序遍历的列表里找出根节点，左边就是左子树，右边就是右子树，中序遍历列表确定左子树和右子树的长度，实际在递归过程中，传递的是左右子树的开始和终止节点

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) {
        return helper(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);
    }
    public TreeNode helper(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd){
        if(preStart>preEnd || inStart>inEnd) return null;
        TreeNode node = new TreeNode(preorder[preStart]);
        int index=0;
        for(int i=inStart; i<=inEnd; i++){
            if(preorder[preStart]==inorder[i]){
                index = i;
                break;
            }
        }
        node.left=helper(preorder, inorder, preStart+1, preStart+index-inStart, inStart, index-1);
        node.right=helper(preorder, inorder, preStart+index-inStart+1, preEnd, index+1, inEnd);
        return node;
        
    }
}