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Leetcode:Symmetric Tree

2024-10-21 14:05:39   208人阅读

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

For example, this binary tree is symmetric:

    1   /   2   2 / \ / 3  4 4  3

 

But the following is not:

    1   /   2   2   \      3    3

 

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

分析：

1.　递归法。如果一个树是symmetric的，那么它的左右子树是镜像对称的。对于判断两个树是否镜像对称，如果两棵树的根节点值相同，并且树A的左孩子跟树B的右孩子镜像对称且树A的右还在和树B的左还在镜像对称，那么树A和B是镜像对称的。代码如下：

 1 class Solution { 2 public: 3     bool isSymmetric(TreeNode *root) { 4         if(root == NULL) return true; 5         return is_mirror(root->left, root->right); 6     } 7     bool is_mirror(TreeNode *l, TreeNode *r){ 8         if(l == NULL || r == NULL) return l == r; 9         if(l->val != r->val) return false;10         return is_mirror(l->left, r->right) && is_mirror(l->right, r->left);11     }12 };

迭代版算法，是利用栈的树迭代遍历算法的变体。主要思想是把对称的两个节点同时Push到栈里，然后每次从栈顶pop两个节点判断是否相同。代码如下：

class Solution {public:    bool isSymmetric(TreeNode *root) {        if(root == NULL) return true;        stack<TreeNode *> s;        s.push(root->left);        s.push(root->right);                while(!s.empty()){            TreeNode *r = s.top();s.pop();            TreeNode *l = s.top();s.pop();            if(r == NULL && l == NULL) continue;//this is very import            if(r == NULL || l == NULL) return false;            if(r->val != l->val) return false;                        s.push(r->left);            s.push(l->right);            s.push(r->right);            s.push(l->left);        }                return true;    }};

 

Leetcode:Symmetric Tree


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