代码随想录算法训练营第二十天 | 654.最大二叉树 | 617.合并二叉树 | 700.二叉搜索树中的搜索 | 98.验证二叉搜索树

最新推荐文章于 2026-01-07 18:44:27 发布

原创最新推荐文章于 2026-01-07 18:44:27 发布 · 93 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法 #leetcode #数据结构

654.最大二叉树

题解及想法

/**
 * 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 constructMaximumBinaryTree1(nums, 0, nums.length);
    }

    private TreeNode constructMaximumBinaryTree1(int[] nums,int leftIndex, int rightIndex){
        if (rightIndex - leftIndex < 1) {// 没有元素了
            return null;
        }
        if (rightIndex - leftIndex == 1) {// 只有一个元素
            return new TreeNode(nums[leftIndex]);
        }
        int maxIndex = leftIndex;// 最大值所在位置
        int maxVal = nums[maxIndex];// 最大值
        for(int i = leftIndex + 1; i < rightIndex; i++){
            if (nums[i] > maxVal){
                maxVal = nums[i];
                maxIndex = i;
            }
        }
        TreeNode root = new TreeNode(maxVal);
        // 根据maxIndex划分左右子树
        root.left = constructMaximumBinaryTree1(nums, leftIndex, maxIndex);
        root.right = constructMaximumBinaryTree1(nums, maxIndex + 1, rightIndex);
        return root;
    }
}

617.合并二叉树

题解及想法

两个树同步递归，把值加起来就行

/**
 * 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) return root2;
        if (root2 == null) return root1;

        root1.val += root2.val;
        root1.left = mergeTrees(root1.left,root2.left);
        root1.right = mergeTrees(root1.right,root2.right);
        return root1;
    }
}

700.二叉搜索树中的搜索

题解及想法

没啥好说的，根据二叉树搜索的特性就行搜索

/**
 * 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 searchBST(TreeNode root, int val) {
        if (root == null || root.val == val) {
                return root;
            }
            if (val < root.val) {  
                return searchBST(root.left, val);
            } else {
                return searchBST(root.right, val);
            }
        }
}

98.验证二叉搜索树

题解及想法

采用中序遍历法，通过一个辅助节点max来保存遍历到的节点，通过对比中间节点root.val 和 max.val的大小进行判断

/**
 * 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 {
    // 递归
    TreeNode max;
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }
        // 左
        boolean left = isValidBST(root.left);
        
        // 中
        if (max != null && root.val <= max.val) {
            return false;
        }
        max = root;
        
        // 右
        boolean right = isValidBST(root.right);

        return left && right;
    }
}