递归题目树型实战

树

实战

226.翻转二叉树

https://leetcode.cn/problems/invert-binary-tree/description/

/**
 * 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 invertTree(TreeNode root) {
        if(root == null) return null;
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26

98.验证二叉搜索树
https://leetcode.cn/problems/validate-binary-search-tree/

/**
 * 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 boolean isValidBST(TreeNode root) {
        return check(root, -(1l << 31), (1l << 31) - 1);
    }

    private boolean check(TreeNode root, long rangeLeft, long rangeRight) {
        if (root == null) return true;
        if (root.val < rangeLeft || root.val > rangeRight) return false;
        return check(root.left, rangeLeft, (long)root.val - 1) && 
            check(root.right, (long)root.val + 1, rangeRight);
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

重叠子问题:翻转or验证左、右子树
当前层逻辑:翻转or验证大小关系
递归边界:叶子节点（无子树)

104.二叉树的最大深度
https://leetcode.cn/problems/maximum-depth-of-binary-tree/

/**
 * 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 int maxDepth(TreeNode root) {
        if(root == null) return 0;
        return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root) {
        ans = 0;
        depth = 1;
        calc(root);
        return ans;
    }
private:
    void calc(TreeNode* root) {
        if (root == nullptr) return ;
        ans = max(ans, depth);
        depth++;
        calc(root->left);
        calc(root->right);
        depth--;
    }
    int depth;
    int ans;
};
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31

111.二叉树的最小深度
https://leetcode.cn/problems/minimum-depth-of-binary-tree/

class Solution {
public:
    int minDepth(TreeNode *root) {
        if (root == nullptr) {
            return 0;
        }

        queue<pair<TreeNode *, int> > que;
        que.emplace(root, 1);
        while (!que.empty()) {
            TreeNode *node = que.front().first;
            int depth = que.front().second;
            que.pop();
            if (node->left == nullptr && node->right == nullptr) {
                return depth;
            }
            if (node->left != nullptr) {
                que.emplace(node->left, depth + 1);
            }
            if (node->right != nullptr) {
                que.emplace(node->right, depth + 1);
            }
        }

        return 0;
    }
};

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28

思路一（自底向上统计信息，分治思想)
最大深度=max(左子树最大深度，右子树最大深度)+1

思路二（自顶向下维护信息)
把“深度”作为一个全局变量——一个跟随结点移动而动态变化的信息递归一层，变量+1，在叶子处更新答案
这种写法需要注意保护和还原现场

推荐一个零声学院免费公开课程，个人觉得老师讲得不错，分享给大家：Linux，Nginx，ZeroMQ，MySQL，Redis，fastdfs，MongoDB，ZK，流媒体，CDN，P2P，K8S，Docker，TCP/IP，协程，DPDK等技术内容，立即学习