【leetcode】100-Same Tree

problem

Same Tree
code

回归的方法

/**
 * 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) );      
    }
};

 or just one line code

class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        return (p==NULL||q==NULL) ? (p==q) : ((p->val==q->val) && isSameTree(p->left, q->left) && isSameTree(p->right, q->right));
    }
};

 这道题还有非递归的解法，因为二叉树的四种遍历(层序，先序，中序，后序)均有各自的迭代和递归的写法，这里我们先来看先序的迭代写法，相当于同时遍历两个数，然后每个节点都进行比较.

参考

1. Same Tree;

2. GrandYang_cnblogs;
3. leeetcode_discuss;
完