【一天一道LeetCode】#101. Symmetric Tree

一天一道LeetCode

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（一）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
Note:
Bonus points if you could solve it both recursively and iteratively.

（二）解题

本题大意：判断一颗树是不是左右对称
解题思路一：分别求二叉树的前序和后序遍历，他们是互为反转数组。

/** * 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 isSymmetric(TreeNode* root) {
            vector<int> pre;
            vector<int> post;
            preOrderTree(root , pre);
            postOrderTree(root , post);
            for(int i = 0 ,j=post.size()-1; i<pre.size(),j>=0;i++,j--)//一个正向，一个反向
            {
                if(pre[i] != post[j]) return false;//不相等就返回false
            }
            return true;
        }
        void preOrderTree(TreeNode* root , vector<int> &pre)//前序遍历
        {
            if(root==NULL) {pre.push_back(0);return;}
            pre.push_back(root->val);
            preOrderTree(root->left,pre);
            preOrderTree(root->right,pre);
        }
        void postOrderTree(TreeNode* root , vector<int> &post)//后续遍历
        {
            if(root==NULL)  {post.push_back(0);return;}
            postOrderTree(root->left,post);
            postOrderTree(root->right,post);
            post.push_back(root->val);
        }
};

解题思路二：将前序遍历和后序遍历合起来一起考虑，对于一个节点，如果他们相等，就判断左右子树是否为镜像！

/** * 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 isSymmetric(TreeNode* root) {
        return isMirror(root,root);
    }
    bool isMirror(TreeNode* root1,TreeNode* root2)
    {
        if(root1==NULL) return root2==NULL;
        if(root2==NULL) return false;
        if(root1->val!=root2->val) return false;
        return isMirror(root1->left,root2->right) && isMirror(root1->right,root2->left);//左子树和右子树是否成镜像
    }
};