[leetCode]100.相同的树

解法一 递归

从根结点开始,首先判断两棵树根节点是否为null,再判断是否相等,然后再对左子树与右子树进行相同的操作。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
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;
        else return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    }
}

解法二 迭代

使用两个双端队列,从根节点开始每次迭代弹出两个对应的节点,
用check方法判断两个节点p、q是否满足:

  • p、q不为null
  • p.val = q.val

如果满足则继续压入子节点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if(p == null && q == null) return true;
        if(!check(p,q)) return false; 

        ArrayDeque<TreeNode> deqP = new ArrayDeque<>();
        ArrayDeque<TreeNode> deqQ = new ArrayDeque<>();
        deqP.addLast(p);
        deqQ.addLast(q);

        while(!deqP.isEmpty()){
            p = deqP.removeFirst();
            q = deqQ.removeFirst();
            if(!check(p,q)) return false;
            if(p!=null){
                if(!check(p.left,q.left)) return false;
                if(p.left != null){//check返回true有两种情况,确保p.left 与 q.left 不为null
                    deqP.addLast(p.left);
                    deqQ.addLast(q.left);
                }
                if(!check(p.right,q.right)) return false;
                if(p.right != null){
                    deqP.addLast(p.right);
                    deqQ.addLast(q.right);
                }
            }
        }
        return true;
    }
    
    //检查pq节点是否相等
    public boolean check(TreeNode p, TreeNode q){
        if(q == null && p == null) return true;
        if(q == null || p == null) return false;
        if(p.val != q.val) return false;
        return true;
    }
}
原文地址:https://www.cnblogs.com/PythonFCG/p/13860015.html