文章目录

• 力扣算法学习day15-1
• • 654-最大二叉树
  • • 题目
    • 代码实现
  • 617-合并二叉树
  • • 题目
    • 代码实现
    • • 已复习 24-两两交换链表中的结点

力扣算法学习day15-1

654-最大二叉树

题目

代码实现

/**
 * 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 constructMaximumBinaryTree(int[] nums) {
        return treeBuild(0,nums.length-1,nums);
    }
    public TreeNode treeBuild(int left,int right,int[] nums){
        // 终止条件
        if(left > right){
            return null;
        }
        if(left == right){// 可以不加这个，加上速度更快，2ms->1ms
            return new TreeNode(nums[left]);
        }
        // 寻找最大值，并记录最大值的坐标。
        int max = nums[left];
        int maxIndex = left;
        for(int i = left;i <= right;i++){
            if(max < nums[i]){
                max = nums[i];
                maxIndex = i;
            }
        }
        // 找到根结点
        TreeNode root = new TreeNode(max);

        // 递归左区间和右区间
        TreeNode leftNode = treeBuild(left,maxIndex-1,nums);
        TreeNode rightNode = treeBuild(maxIndex+1,right,nums);

        // 连接左右结点
        root.left = leftNode;
        root.right = rightNode;

        return root;
    }
}

617-合并二叉树

题目

代码实现

/**
 * 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 mergeTrees(TreeNode root1, TreeNode root2) {
        if(root1 == null && root2 != null){
            return root2;
        } else if(root2 == null && root1 != null){
            return root1;
        } else if(root1 == null && root2 == null){
            return null;
        }
        root1.val = root1.val + root2.val;
        root1.left = mergeTrees(root1.left,root2.left);
        root1.right = mergeTrees(root1.right,root2.right);

        return root1;
    }
}

已复习 24-两两交换链表中的结点

标签：right,TreeNode,val,int,力扣,day15,算法,root1,left
来源： https://blog.csdn.net/weixin_45609749/article/details/122782143