LeetCode 100 Same Tree

Problem:

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

Summary:

判断两棵二叉树是否完全相同。

Analysis:

1. 判断两棵树是否完全相同，则需遍历每个节点，最容易想到的写法为递归。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        if (!p && !q) {
            return true;
        }
        else if (!p || !q || p->val != q->val) {
            return false;
        }
        
        return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
    }
};

 2. 非递归写法，利用queue对两树进行层次遍历，分别比较每个节点。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        queue<TreeNode*> q1, q2;
        q1.push(p); q2.push(q);
        
        while (!q1.empty() && !q2.empty()) {
            TreeNode* tmp1 = q1.front(); 
            TreeNode* tmp2 = q2.front(); 
            q1.pop(); q2.pop();
            
            if (!tmp1 ^ !tmp2 || (tmp1 && tmp2 && tmp1->val != tmp2->val)) {
                return false;
            }
            
            if (tmp1) {
                q1.push(tmp1->left);
                q1.push(tmp1->right);
            }
            
            if (tmp2) {
                q2.push(tmp2->left);
                q2.push(tmp2->right);
            }
            
            if (q1.size() != q2.size()) {
                return false;
            }
        }
        
        return true;
    }
};