[leetcode]: 101. Symmetric Tree

1.题目

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    3

给一棵二叉树，判断这棵树是否镜像对称。

2.分析

广度优先遍历。
左子树和右子树遍历的方向相反即可。
递归or迭代

3.代码

递归

class Solution {public:    bool judgeSysmetric(TreeNode* left, TreeNode*right) {        if (left == NULL || right == NULL)            return left == right;        if (left->val!=right->val)            return false;        return judgeSysmetric(left->left, right->right) && judgeSysmetric(left->right, right->left);    }    bool isSymmetric(TreeNode* root) {        if (root == NULL)            return true;        return judgeSysmetric(root->left, root->right);    }};

迭代

bool isSymmetric(TreeNode* root) {    if (root == NULL)        return true;    //左右子树分别用queue存储    queue<TreeNode*> left;    left.push(root->left);    queue<TreeNode*> right;    right.push(root->right);    while (!left.empty() && !right.empty()) {        TreeNode* l = left.front();        left.pop();        TreeNode* r = right.front();        right.pop();        if (l == NULL && r == NULL)            continue;        if (l == NULL || r == NULL)            return false;        if (l->val != r->val)            return false;        //左边与右边遍历的方向相反。左边从左到右。右边从右到左。        left.push(l->left);        left.push(l->right);        right.push(r->right);        right.push(r->left);    }    return true;}