101. Symmetric Tree

问题描述

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.

题目链接：

思路分析

给出一棵二叉树，判断这颗二叉树是否是对称的。对称的条件是以root所在的位置为对称轴，左右子树的结构和值都相同。一开始想用100. Same Tree的中序遍历树然后对比头尾是否相同的方式解决， 发现这样做会有一些结构上的不同无法发现。

然后现在使用的是递归的方式，检查每一个点是否相同，然后在比较左子结点的左子树与右子结点的右子树，左子结点的右子树与右子结点的左子树是否相同就可以了。

代码

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root != null){
            return compare(root.left, root.right);
        }
        else{
            return true;
        }
    }
    
    private boolean compare(TreeNode p, TreeNode q){
        if (p == null && q == null){
            return true;
        }
        else if (p == null || q == null){
            return false;
        }
        else if (p.val != q.val){
            return false;
        }
        else{
            return compare(p.left, q.right) && compare(p.right, q.left);
        }
    }
}

/**
 * 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) {
        if(root)
            return compare(root->left, root->right);
        else
            return true;
            
    }
    
    bool compare(TreeNode* p, TreeNode* q){
        if (!p && !q)
            return true;
        else if (!p || !q)
            return false;
        else if (p->val != q->val)
            return false;
        else
            return (compare(p->left, q->right) && compare(p->right, q->left));
    }
};

时间复杂度：O(n) // n为树的结点数

---

####反思
还是没有掌握递归的精髓，不要总是想着把所有结点都处理完再进行比较，错误代码：

/**
 * 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) {
        string s = inorderVisitTree(root);
        int n = s.length();
        for(int i = 0; i < (n/2 + 1); i++){
            if(s[i] != s[n-1-i])
                return false;
        }
        return true;
    }
    
     string inorderVisitTree(TreeNode* p){
        string s = "";
        if(p){
            if (p->left == NULL && p->right == NULL)
                return s += p->val;
            if (p->left != NULL)
                s += inorderVisitTree(p->left);
            else
                s += "n";
            s += p->val;
            if(p->right != NULL)
                s += inorderVisitTree(p->right);
            else
                s += "n";
        }
        else
            return "";
        return s;
    }
    
};

。