[每日 LeetCode] 100. Same Tree

原文链接 [每日 LeetCode] 100. Same Tree

Description:

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.

Example 1:

Input:     
	   1         1
          / \       / \
         2   3     2   3

        [1,2,3],   [1,2,3]

Output: true

Example 2:

Input:
          1         1
          /           \
         2             2

        [1,2],     [1,null,2]

Output: false

Example 3:

Input:
           1         1
          / \       / \
         2   1     1   2

        [1,2,1],   [1,1,2]

Output: false

思路：本题判断两个树是否是同一棵，同样可以采用递归和非递归方法实现。递归方法比较简洁，分别对当前结点，左结点和右结点依次比较即可。非递归可采用层次、先序、中序和后序遍历的任何一种，这里使用中序遍历的非递归写法，遍历后判断相应位置上的结点是否相同即可。

C++ 代码（递归方法）

/**
 * 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 (p == q);
    return (p->val == q->val && isSameTree(p->left, q->left) && isSameTree(p->right, q->right));
    }
};

运行时间：4ms

运行内存：10M

C++ 代码（非递归方法）

/**
 * 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) {
        stack<TreeNode*> pcallStack;
        stack<TreeNode*> qcallStack;
        
        TreeNode* pit = p;
        TreeNode* qit = q;
        
        if((!pit && qit) || (pit && !qit))
            return false;
        
        while((pit && qit) || (!pcallStack.empty() && !pcallStack.empty()) ) {
            if((!pit && qit) || (pit && !qit))
                return false;
            
            if(pit && qit) {
                pcallStack.push(pit);
                pit=pit->left;
                
                qcallStack.push(qit);
                qit=qit->left;
            } else {
                if(pcallStack.top()->val != qcallStack.top()->val)
                    return false;
                    
                pit = pcallStack.top()->right;
                pcallStack.pop();
                qit = qcallStack.top()->right;
                qcallStack.pop();
            }
        }
        
        return true;
    }
};

运行时间：4ms

运行内存：10.1M