leetcode 101. Symmetric Tree

101. Symmetric Tree对称树判断c语言实现

Description

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1   / \  2   2 / \ / \3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1   / \  2   2   \  /   3 2

Note:
Bonus points if you could solve it both recursively and iteratively.（使用循环或递归是加分点）

思路：这道题其实可以使用中序遍历来进行判断，但由于c语言对于动态数组的处理较繁琐，不像C++有vector，java有list。下面给出一种递归求法，c++版的中序遍历方法可以见博客http://blog.csdn.net/jin_kwok/article/details/51162625

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     struct TreeNode *left; *     struct TreeNode *right; * }; */bool checkSon(struct TreeNode *leftSon, struct TreeNode *rightSon){    if(leftSon == NULL && rightSon == NULL) return 1; //没有孩子节点    else if(leftSon == NULL || rightSon == NULL) return 0; //只有左/右孩子    else if(leftSon->val == rightSon->val) //左右孩子存在且值相等    {        return checkSon(leftSon->left, rightSon->right) && checkSon(leftSon->right, rightSon->left);    }    else return 0; //左右孩子存在且值不相等}bool isSymmetric(struct TreeNode* root) {    if(root == NULL)  return 1;  //空树    else return checkSon(root->left,root->right);}