[LeetCode] Same Tree

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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.

解题思路：

递归做即可。两颗二叉树相同，当且仅当当前节点值相同，并且对应的左右孩子树也相同。

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/**
 * 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){   //其中有一个是空
            return false;
        }
        if(p->val!=q->val){
            return false;
        }
        return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
    }
};

另外一种非递归法，用队列来记录两棵树当前状态：

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/**
 * 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){   //其中有一个是空
            return false;
        }
        //逻辑上，两个队列的大小相同
        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; } };
   
  

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