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.
递归的解法：
用树递归的思想，将两个树以结点作为比较单位，只关注对当前位置结点的操作（是否都存在此结点，存在时值是否相同）。
注意：
1、只关注单个结点，结点的孩子交给递归去做，这样可以最简化逻辑；
2、结点是否存在、结点值是否相等、结点是否有孩子，是三个不同的概念（“结点是否有孩子”的概念本题中未体现）；
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 11 class Solution { 12 public: 13 bool isSameTree(TreeNode *p, TreeNode *q) { 14 if (!p && !q) 15 return true; 16 17 if ((p && !q) || (!p && q) || (p->val != q->val)) 18 return false; 19 20 return isSameTree(p->left, q->left) && isSameTree(p->right, q->right); 21 } 22 };
迭代的解法：
引用先序/中序/后序/按层等遍历二叉树的思想，用堆栈（数组）存放遍历到的结点，进栈出栈的同时进行比较。
中序遍历（其他方法类似）：
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isSameTree(TreeNode *p, TreeNode *q) { 13 vector<TreeNode *> stackP; 14 vector<TreeNode *> stackQ; 15 TreeNode *currentP = p; 16 TreeNode *currentQ = q; 17 18 while (currentP || stackP.size()) { 19 if (!currentP && !currentQ) { 20 currentP = stackP.back(); 21 currentQ = stackQ.back(); 22 stackP.pop_back(); 23 stackQ.pop_back(); 24 currentP = currentP->right; 25 currentQ = currentQ->right; 26 continue; 27 } 28 29 if ((!currentP && currentQ) || (currentP && !currentQ) || 30 (currentP->val != currentQ->val)) 31 return false; 32 33 if (currentP->val == currentQ->val) { 34 stackP.push_back(currentP); 35 stackQ.push_back(currentQ); 36 currentP = currentP->left; 37 currentQ = currentQ->left; 38 } 39 } 40 41 if (!currentQ) { 42 return true; 43 } else { 44 return false; 45 } 46 47 } 48 };
附录：
C++中vector、stack、queue用法和区别
递归/非递归遍历二叉树