LeetCode ---- 572. 另一个树的子树 （迭代/递归， java，c++）

判断二叉树是否是另一棵二叉树的子树

方法一：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        //思路：将两棵树序列化成一个String类型，判断s1是否包含s2即可
        //序列化规则：空子树用#代替，结点分隔符用！
        String s1 = preOrder(s);
        String s2 = preOrder(t);
    
        if(s1.contains(s2)){
            return true;
        }else{
            return false;
        }
    }
    //迭代：先序遍历
    public String preOrder(TreeNode tr){
        //注意一开始也要有结点分隔符，不然[12]和[2]这两棵树会被判断为true
        StringBuilder sb = new StringBuilder("!");
        Stack stack = new Stack();
        TreeNode t = tr;
        while(true){
            while(t != null){
                sb.append(String.valueOf(t.val));
                 sb.append("!"); //结点分隔符用！
                stack.push(t);
                t = t.left;
            }
            sb.append("#!"); //空子树用#，结点分隔符用！
            if(!stack.isEmpty()){
                TreeNode top = stack.pop();
                t = top.right;
            }else{
                break;
            }
        }
        return sb.toString();  
    }
   
}

 

方法二：

思路

遍历整个s树，看以s树的每个结点为根结点的子树是不是与目标子树t一致，若有一致的，返回true，否则返回false

 

方法二解法1，层次遍历s

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        //层次遍历s,依次比较【以s中的结点为根结点的子树】与t是否相等
        Queue q = new LinkedList();
        q.offer(s);
        while(!q.isEmpty()){
            TreeNode ts = q.poll();
            boolean res = sameTree(ts,t);
            if(res == true){
                return true;
            }
            
            if(ts.left != null){
                q.offer(ts.left);
            }
            if(ts.right != null){
                q.offer(ts.right);
            }
        }    
        return false;
        
    }
   
    public boolean sameTree(TreeNode s, TreeNode t){
        if(s == null && t == null){
            return true;
        }
        if(s == null || t == null){
            return false;
        }
        if(s.val != t.val){
            return false;
        }
        
        return sameTree(s.left, t.left) && sameTree(s.right, t.right);
    }
}

 

方法二解法2，递归调用：

 

c++实现:

bool isSubtree(TreeNode* s, TreeNode* t) {
    if (!s) return false;
    if (sameTree(s, t)) return true;
    return isSubtree(s->left, t) || isSubtree(s->right, t);
}

bool sameTree(TreeNode* s, TreeNode* t) {
    if (!s && !t) return true;
    if (!s || !t) return false;
    if (s->val != t->val) return false;
    return sameTree(s->left, t->left) && sameTree(s->right, t->right);
}