[LeetCode] Same Tree

[LeetCode] Same Tree

Same Tree

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.

解題思路：

遞歸做即可。兩顆二叉樹相同，當且僅當當前節點值相同，並且對應的左右孩子樹也相同。

/**
 * 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==NULL && q==NULL){
            return true;
        }else if(p==NULL || q==NULL){   //其中有一個是空
            return false;
        }
        if(p->val!=q->val){
            return false;
        }
        return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
    }
};

另外一種非遞歸法，用隊列來記錄兩棵樹當前狀態：

/**
 * 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==NULL && q==NULL){
            return true;
        }else if(p==NULL || q==NULL){   //其中有一個是空
            return false;
        }
        //邏輯上，兩個隊列的大小相同
        queue queue1({p});
        queue queue2({q});
        while(!queue1.empty()){
            TreeNode* node1 = queue1.front();
            TreeNode* node2 = queue2.front();
            queue1.pop();
            queue2.pop();
            if(node1->val!=node2->val){
                return false;
            }
            if(node1->left!=NULL && node2->left!=NULL){
                queue1.push(node1->left);
                queue2.push(node2->left);
            }else if(node1->left!=NULL || node2->left!=NULL){
                return false;
            }
            if(node1->right!=NULL && node2->right!=NULL){
                queue1.push(node1->right);
                queue2.push(node2->right);
            }else if(node1->right!=NULL || node2->right!=NULL){
                return false;
            }
        }
        return true;
    }
};