https://www.acwing.com/problem/content/39/
直接模拟

```java
class Solution {
    public int[] printMatrix(int[][] matrix) {
        if(matrix == null || matrix .length == 0 || matrix[0].length == 0) {
            return new int[0];
        }
        
        int row = matrix.length;
        
        int col = matrix[0].length;
        
        int top = 0;
        int bot = row - 1;
        int left = 0;
        int right = col - 1;
        
        int index = 0;
        
        
        int[] result = new int[row * col];
        
        while(left <= right && top <= bot) {
            for(int i = left; i <= right; i++) {
                result[index++] = matrix[top][i]; 
            }
            
            top++;
            
            for(int i = top; i <= bot; i++) {
                result[index++] = matrix[i][right];
            }
            
            right--;
            
            
            if(top <= bot) {
                for(int i = right; i >= left; i--) {
                    result[index++] = matrix[bot][i];
                }
                
                bot--;
            }
            
            if(left <= right) {
                for(int i = bot; i >= top; i--) {
                    result[index++] = matrix[i][left];
                }
                
                left++;
            }
            
            
        }
        
        return result;
        
    }
}

```

9.11算法题

https://www.acwing.com/problem/content/35/


今天做这题

```java
class Solution {
    public boolean hasSubtree(TreeNode pRoot1, TreeNode pRoot2) {
         if(pRoot1 == null || pRoot2 == null) {
            return false;
        }
        
        return v(pRoot1,pRoot2) || hasSubtree(pRoot1.left,pRoot2) || hasSubtree(pRoot1.right,pRoot2);
    }
    
    private boolean v(TreeNode A,TreeNode B) {
        if(B == null) {
            return true;
        }
        
        if(A == null || A.val != B.val){
            return false;
        }
        
        return v(A.left,B.left) && v(A.right, B.right);
    }
}
```


```go
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func hasSubtree(pRoot1 *TreeNode, pRoot2 *TreeNode) bool {
    if(pRoot2 == nil || pRoot1 == nil) {
        return false;
    }
    
    return V(pRoot1, pRoot2) || hasSubtree(pRoot1.Left, pRoot2) || hasSubtree(pRoot1.Right, pRoot2);
}


func V(A, B *TreeNode) bool {
    if(B == nil) {
        return true;
    }
    
    if(A == nil || A.Val != B.Val) {
        return false;
    }
    
    return V(A.Left, B.Left) && V(A.Right, B.Right);
}
```

9.12算法题

今天写这题

https://www.acwing.com/problem/content/submission/code_detail/36912299/

```java

class Solution {
    public boolean verifySequenceOfBST(int [] sequence) {
        if (sequence == null || sequence.length == 0) {
            return true;
        }
        return verify(sequence, 0, sequence.length - 1);
    }

    private boolean verify(int[] postorder, int start, int end) {
        if (start >= end) {
            return true;
        }

        int root = postorder[end];
        int i = start;

        // 找到第一个大于根节点的元素，这个元素之前的部分是左子树
        while (i < end && postorder[i] < root) {
            i++;
        }

        // 检查右子树中的所有元素是否都大于根节点
        for (int j = i; j < end; j++) {
            if (postorder[j] < root) {
                return false;
            }
        }

        // 递归检查左子树和右子树
        return verify(postorder, start, i - 1) && verify(postorder, i, end - 1);
    }

}

```



9.10算法题

2012.12阅读精读

https://www.acwing.com/problem/content/43/

今天写这题，bfs层序遍历去遍历，代码如下

```java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> printFromTopToBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        
        if (root == null) {
            return res;
        }
        
        
        Queue<TreeNode> queue = new LinkedList<>();
        Boolean isLeft = true;
        
        queue.offer(root);
        while(!queue.isEmpty()){
            
            List<Integer> currentLevel = new ArrayList<>();
            
            Integer size = queue.size();
            
            for(int i = 0;i < size; i++) {
                
                TreeNode node = queue.poll();
                currentLevel.add(node.val);
                
                if(node.left != null) {
                    queue.add(node.left);
                }
                
                if(node.right != null) {
                    queue.add(node.right);
                }
                
                
            }
            
            
            if(!isLeft) {
                 Collections.reverse(currentLevel);
            }
            
            res.add(currentLevel);
            isLeft = !isLeft;
            
            
        }
        
        return res;
    }
}
```

短篇1