等价二叉树-lintcode

看两个二叉树形状和内容是否完全相等。可以递归求解，每一个调用栈内存储两个返回的bool结果：lflag和rflag。然后返回它们的与。同样，和通常的递归思想一样，调用栈开始就要做好递归结束的判断：a、b都为NULL返回true，一个为NULL另一个不为NULL返回false，val不相同返回false。

C++代码：

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {public:    /**     * @aaram a, b, the root of binary trees.     * @return true if they are identical, or false.     */    bool isIdentical(TreeNode* a, TreeNode* b) {        return search(a, b);    }    bool search(TreeNode * a, TreeNode * b){        if (a == NULL && b == NULL) {            return true;        }        if (a == NULL || b == NULL) {            return false;        }        if (a->val != b->val) {            return false;        }        bool lflag = search(a->left, b->left);        bool rflag = search(a->right, b->right);        return (lflag && rflag);            }};

Python代码：

"""Definition of TreeNode:class TreeNode:    def __init__(self, val):        this.val = val        this.left, this.right = None, None"""class Solution:    """    @param a, b, the root of binary trees.    @return true if they are identical, or false.    """    def isIdentical(self, a, b):        return self.search(a,b)    def search(self,a,b):        if a==None and b==None:            return True        if a==None or b==None:            return False        if a.val != b.val:            return False        a1 = self.search(a.left, b.left)        a2 = self.search(a.right, b.right)        return a1 and a2