leetCode(15):Symmetric Tree

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.

首先递归方法：

bool isChildSymmectric(TreeNode* node1,TreeNode* node2){if(node1==NULL && node2==NULL)return true;if(node1==NULL || node2==NULL || node1->val!=node2->val)return false;return isChildSymmectric(node1->left,node2->right) && isChildSymmectric(node1->right,node2->left);}bool isSymmetric(TreeNode* root){if(root==NULL)return true;return isChildSymmectric(root->left,root->right);}

循环方法：把左右子树依次压入到容器中，并一层一层地判断，如果不平衡则立即返回；直至没有新的结点压入到容器中。

/** * 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==NULL)    return true;    vector<TreeNode*> nodes;    nodes.push_back(root);    int nodesBegin=0;    int nodesEnd=nodes.size();        while(nodesBegin<nodesEnd)    {    int b=nodesBegin;    int e=nodes.size()-1;    while(b<e)    {    if(nodes[b]==NULL && nodes[e]==NULL)    {    b++;    e--;    continue;    }    if((nodes[b]==NULL && nodes[e]) || (nodes[b] && nodes[e]==NULL)    || (nodes[b]->val!=nodes[e]->val))    return false;        nodes.push_back(nodes[b]->left);    nodes.push_back(nodes[b]->right);        b++;    e--;    }    while(b<nodesEnd)    {    if(NULL==nodes[b])    {    b++;    continue;    }    nodes.push_back(nodes[b]->left);    nodes.push_back(nodes[b]->right);    b++;    }    nodesBegin=nodesEnd;    nodesEnd=nodes.size();    }    return true;    }};