100. Same Tree

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)。

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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.
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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;
  }
}
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