LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal 二叉树构建（dfs）

dfs每次规定前序和中序的边界，递归构建左右儿子

注意预处理每个序号在中序里的位置，以免超时

/**
 * 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 {
    int []pre = new int [3010];
    int []in  = new int [3010];
    int []pos = new int[6010];
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length == 0)
            return null;
        pre = preorder;
        in = inorder;
        int r_pre = preorder.length;
        int r_in  = r_pre;
        int l_in  = 0;
        int l_pre = 0;
        for(int i=0;i<r_pre;i++)
            pos[in[i]+3000] = i;
        TreeNode head = createNode(l_pre,r_pre,l_in,r_in);
        return head;
    }
    public TreeNode createNode(int l_pre, int r_pre, int l_in, int r_in){
        // System.out.println(l_pre);
        // System.out.println(r_pre);
        // System.out.println(l_in);
        // System.out.println(r_in);
        // System.out.println();
        if(l_pre >= r_pre)
            return null;
        TreeNode ans = new TreeNode(pre[l_pre]);
        int mid_pos  = pos[pre[l_pre]+3000];
        int num_left = mid_pos - l_in;
        int num_rig  = r_in - mid_pos-1;
        ans.left = createNode(l_pre+1,l_pre+1+num_left,l_in,mid_pos);
        ans.right = createNode(r_pre - num_rig,r_pre,mid_pos+1,r_in);
        return ans;
    }
}