力扣labuladong——一刷day71

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前言

二叉树的递归分为「遍历」和「分解问题」两种思维模式，这道题需要用到「分解问题」的思维模式。

一、力扣1008. 前序遍历构造二叉搜索树

/**
 * 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 {
    Map<Integer,Integer> map = new HashMap<>();
    public TreeNode bstFromPreorder(int[] preorder) {
        int[] inorder = new int[preorder.length];
        System.arraycopy(preorder,0,inorder,0,preorder.length);
        Arrays.sort(inorder);
        for(int i = 0; i < inorder.length; i ++){
            map.put(inorder[i],i);
        }
        return fun(preorder,inorder,0,preorder.length-1,0,inorder.length-1);
    }
    public TreeNode fun(int[] preorder, int[] inorder, int proS, int proE, int inS, int inE){
        if(proS > proE || inS > inE){
            return null;
        }
        TreeNode cur = new TreeNode(preorder[proS]);
        int index = map.get(preorder[proS]);
        int len = index - inS;
        cur.left = fun(preorder,inorder,proS+1,proS+len,inS, index-1);
        cur.right = fun(preorder,inorder,proS+len+1,proE,index+1,inE);
        return cur;
    }
}

二、力扣108. 将有序数组转换为二叉搜索树

/**
 * 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 {
    public TreeNode sortedArrayToBST(int[] nums) {
        return fun(nums,0,nums.length-1);
    }
    public TreeNode fun(int[] nums, int low, int high){
        if(low > high){
            return null;
        }
        int mid = (high+low)/2;
        TreeNode cur = new TreeNode(nums[mid]);
        cur.left = fun(nums,low,mid-1);
        cur.right = fun(nums,mid+1,high);
        return cur;
    }
}