101. Symmetric Tree (Tree, Queue; DFS, WFS)

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

For example, this binary tree is symmetric:

But the following is not:

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

思路：要等到左儿子和右儿子的结果都知道了，才能判断当前节点的对称性，所以是后序遍历

法I：递归后序遍历

class Solution {

public:

    bool isSymmetric(TreeNode *root) {

        if(!root) return true;

        bool result;

        if((root->left==NULL && root->right != NULL) ||(root->left!=NULL && root->right == NULL))

        {

            return false;

        }

        else if(root->left == NULL && root->right == NULL)

        {

            return true;

        }

        else

        {

            result = cmp(root->left, root->right);

        }

    }

    bool cmp(TreeNode * node1, TreeNode* node2)

    {

        int result1 = true;

        int result2 = true;

        if(node1->val!=node2->val) return false;

        else

        {

            //递归结束条件：至少有一个节点为NULL

            if((node1->left==NULL && node2->right != NULL) ||

            (node1->left!=NULL && node2->right == NULL)||

            (node1->right!=NULL && node2->left == NULL)||

            (node1->right==NULL && node2->left != NULL))

            {

                return false;

            }

            if((node1->left == NULL && node2->right == NULL)&&

            (node1->right == NULL && node2->left== NULL))

            {

                return true;

            }

            //互相比较的两个点，要比较节点1的左儿子和节点2的右儿子，以及节点1的右儿子和节点2的左儿子

            if(node1->left != NULL)

            {

                result1 = cmp(node1->left,node2->right);

            }

            if(node1->right != NULL)

            {

                result2 = cmp(node1->right,node2->left);

            }

            return (result1 && result2);

        }

    }

};

法II：用队列实现层次遍历（层次遍历总是用队列来实现）

class Solution {

public:

    bool isSymmetric(TreeNode *root) {

        if(root == NULL)    return true;

        queue<TreeNode*> q;

        q.push(root->left);

        q.push(root->right);

        TreeNode *t1, *t2;

        while(!q.empty()){

            t1 = q.front();

            q.pop();

            t2 = q.front();

            q.pop();

            if(t1 == NULL && t2 == NULL)

                continue;

            if(t1 == NULL || t2 == NULL || t1->val != t2->val)

                return false;

            q.push(t1->left);

            q.push(t2->right);

            q.push(t1->right);

            q.push(t2->left);

        }

        return true;

    }

};

巴特西

101. Symmetric Tree (Tree, Queue; DFS, WFS)

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