代码随想录训练营Day17

654.最大二叉树 leetcode.com/problems/ma…

思路：构造二叉树题目

class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        return constructMaximumBinaryTree1(nums, 0, nums.length);
    }

    public 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 leetcode.com/problems/me…

思路：比较简单，递归就行

class Solution {
    // 递归
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
    // 同时操作两个节点
        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 leetcode.com/problems/se…

思路：二叉搜索树比普通二叉树多了特性

class Solution {
    // 递归，普通二叉树
    public TreeNode searchBST(TreeNode root, int val) {
        if (root == null || root.val == val) {
            return root;
        }
        TreeNode left = searchBST(root.left, val);
        if (left != null) {
            return left;
        }
        return searchBST(root.right, val);
    }
}

98.验证二叉搜索树 leetcode.com/problems/va…

思路：中序遍历下，输出的二叉搜索树节点的数值是有序序列。验证二叉搜索树，就相当于变成了判断一个序列是不是递增的了

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