相同的树

二叉树的遍历

前序遍历：

/**
 * 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 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  isSameTree(p.left,q.left) && isSameTree(p.right,q.right); 
        }else{
            return false;
        }
    }
}

层次遍历

/**
 * 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 isSameTree(TreeNode p, TreeNode q) {
        Queue<TreeNode> s1 = new LinkedList<>();
        Queue<TreeNode> s2 = new LinkedList<>();
        if(p == null && q == null) return true;
        if(p == null || q ==null) return false;
        s1.add(p);s2.add(q);
        while(!s1.isEmpty() && !s2.isEmpty()){
            int size1 = s1.size();
            for(int i=0;i<size1;i++){
              TreeNode t1 = s1.poll();
              TreeNode t2 = s2.poll();
              if(t1.val != t2.val) return false;
              //下面的if判断可以保证两个队列的大小一定相等否则就直接返回false了    
              if(t1.left != null && t2.left != null){
                  s1.add(t1.left);
                  s2.add(t2.left);
              }
             if(t1.right != null && t2.right != null){
                  s1.add(t1.right);
                  s2.add(t2.right);
              }
              if(t1.left != null ^ t2.left != null){//异或，当一个为null,一个不为null时为真
                  return false;
              }
              if(t1.right != null ^ t2.right != null){
                  return false;
              }

            }
        }
        return true;
        
    }

}

两个解法时间复杂度都差不多