100. Same Tree

最新推荐文章于 2024-10-14 18:27:49 发布

原创最新推荐文章于 2024-10-14 18:27:49 发布 · 152 阅读

0 ·

CC 4.0 BY-SA版权

Tree 同时被 2 个专栏收录

84 篇文章

订阅专栏

温故知新

53 篇文章

订阅专栏

博客围绕判断两棵二叉树是否相同展开。题目要求判断两棵二叉树结构是否相同且节点值相等。给出两种方法思路，分别是递归法（recursive）和迭代法（iterative）。

题目描述

Given two binary trees, write a function to check if they are the same or not.
Two binary trees are considered the same if they are structurally identical and the nodes have the same value.
在这里插入图片描述

方法思路

Appraoch1:recursive

class Solution {
    //Runtime: 2 ms, faster than 100.00% 
    //Memory Usage: 36.8 MB, less than 73.43%
    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);
    }
}

Appraoch2:iterative

class Solution {
    //Runtime: 2 ms, faster than 100.00%
    //Memory Usage: 37 MB, less than 5.52% 
  public boolean check(TreeNode p, TreeNode q) {
    // p and q are null
    if (p == null && q == null) return true;
    // one of p and q is null
    if (q == null || p == null) return false;
    if (p.val != q.val) return false;
    return true;
  }

  public boolean isSameTree(TreeNode p, TreeNode q) {
    if (p == null && q == null) return true;
    if (!check(p, q)) return false;

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

    while (!deqP.isEmpty()) {
      p = deqP.removeFirst();
      q = deqQ.removeFirst();

      if (!check(p, q)) return false;
      if (p != null) {
        // in Java nulls are not allowed in Deque
        if (!check(p.left, q.left)) return false;
        if (p.left != 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;
  }
}