二叉树习题（上）

Posted 2022-12-15 *平芜尽处是春山*

下面是总集呐
100.相同的树
572.另一棵树的子树
110. 平衡二叉树
101.对称二叉树

100.相同的树

class Solution 
    public boolean isSameTree(TreeNode p, TreeNode q) 
        if(p == null && q == null) 
            return true;
        
        if(p == null || q == null) 
            return false;
        
        if(p.val != q.val) 
            return false;
        
        return isSameTree(p.left,q.left) && isSameTree(p.right,q.right);

    

运行截图：

572.另一棵树的子树

在这里插入代码片class Solution 
    private boolean isSameTree(TreeNode p,TreeNode q)
        if(p==null&&q==null)
            return true;
        
        if(p!=null&&q==null||p==null&&q!=null)
            return false;
        
        if(p.val!=q.val)
            return false;
        
        return isSameTree(p.left,q.left)&&isSameTree(p.right,q.right);
    
    
    public boolean isSubtree(TreeNode root, TreeNode subRoot) 
        if(root==null||subRoot==null)
            return false;
        
        if(isSameTree(root,subRoot))
            return true;
        
        if(isSubtree(root.left,subRoot))
            return true;
        
        if(isSubtree(root.right,subRoot))
            return true;
        
        return false;
    

运行截图：

110. 平衡二叉树

class Solution 
    public boolean isBalanced(TreeNode root) 
        if(root == null) 
            return true;
        
        int leftHeight = height(root.left);
        int rightHeight = height(root.right);
        int heightAbs = Math.abs(leftHeight - rightHeight);
        if(heightAbs < -1 || heightAbs > 1) 
            return false;
        
        return isBalanced(root.left) && isBalanced(root.right);
    

    public int height(TreeNode root) 
        if(root == null)
            return 0;
        
        return 1 + Math.max(height(root.left),height(root.right));
    

运行截图：

代码优化：

class Solution 
    public boolean isBalanced(TreeNode root) 
        if(root == null) 
            return true;
        
        Map<TreeNode,Integer> map = new HashMap<> ();
        int leftHeight = 0;
        int rightHeight = 0;
        if(map.containsKey(root.left)) 
            leftHeight = map.get(root.left);
         else 
            leftHeight = height(root.left);
            map.put(root.left,leftHeight);
        
        if(map.containsKey(root.right)) 
            rightHeight = map.get(root.right);
         else 
            rightHeight = height(root.right);
            map.put(root.right,rightHeight);
        
        int heightAbs = Math.abs(leftHeight - rightHeight);
        if(heightAbs < -1 || heightAbs > 1) 
            return false;
        
        return isBalanced(root.left) && isBalanced(root.right);
    

    public int height(TreeNode root) 
        if(root == null)
            return 0;
        
        return 1 + Math.max(height(root.left),height(root.right));
    

运行截图：

101.对称二叉树

1、普通方法

class Solution 
    public boolean isSymmetric(TreeNode root) 
        if(root == null) 
            return true;
        
        return isMirror(root.left,root.right);
    

    private boolean isMirror(TreeNode left,TreeNode right) 
        if(left == null && right == null) 
            return true;
        
        if(left == null || right == null) 
            return false;
        
        if(left.val != right.val) 
            return false;
        
        return isMirror(left.left,right.right) && isMirror(left.right,right.left);
    

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2、使用队列

class Solution 
    public boolean isSymmetric(TreeNode root) 
        if(root == null || (root.left == null && root.right == null)) 
            return true;
        

        Queue<TreeNode> queue = new LinkedList<> ();
        queue.offer(root.left);
        queue.offer(root.right);
        while(!queue.isEmpty()) 
            TreeNode left = queue.poll();
            TreeNode right = queue.poll();
            if(left == null && right == null) 
                continue;
            
            if(left == null || right == null) 
                return false;
            
            if(left.val != right.val) 
                return false;
            
            queue.offer(left.left);
            queue.offer(right.right);
            queue.offer(left.right);
            queue.offer(right.left);
        
        return true; 
       

运行截图：