38. Same Tree && Symmetric Tree

Same Tree

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

思想： 无。能遍历即可。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSameTree(TreeNode *p, TreeNode *q) {
        if((p == NULL && q) || (p && q == NULL)) return false;
        if(p == NULL && q == NULL) return true;
        if(p->val != q->val) return false;
        return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
    }
};

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.

思想： 构造其镜像树。

1. 递归。用 Same Tree方法判断即可。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
TreeNode* getMirror(TreeNode *root) {
    if(root == NULL) return NULL;
    TreeNode *p = new TreeNode(root->val);
    p->left = getMirror(root->right);
    p->right = getMirror(root->left);
    return p;
}
bool isSymmetricCore(TreeNode *root, TreeNode *root2) {
    if((!root && root2) || (root && !root2)) return false;
    if(!root && !root2) return true;
    if(root->val != root2->val) return false;
    return isSymmetricCore(root->left, root2->left) && isSymmetricCore(root->right, root2->right);
}
class Solution {
public:
    bool isSymmetric(TreeNode *root) {
        TreeNode *mirrorTree = getMirror(root);
        return isSymmetricCore(root, mirrorTree);
    }
};

2. 迭代。两棵树相同方法遍历即可。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
TreeNode* getMirror(TreeNode *root) {
    if(root == NULL) return NULL;
    TreeNode *p = new TreeNode(root->val);
    p->left = getMirror(root->right);
    p->right = getMirror(root->left);
    return p;
}
bool pushChildNode(TreeNode *p1, TreeNode *p2, queue<TreeNode*> & qu, queue<TreeNode*>& qu2) {
    if(p1->left && p2->left) { qu.push(p1->left); qu2.push(p2->left); }
    else if(p1->left || p2->left) return false;
    if(p1->right && p2->right) { qu.push(p1->right); qu2.push(p2->right);}
    else if(p1->right || p2->right) return false;
    return true;
}
class Solution {
public:
    bool isSymmetric(TreeNode *root) {
        if(root == NULL) return true;
        TreeNode *mirrorTree = getMirror(root);
        queue<TreeNode*> que1;
        queue<TreeNode*> que2;
        que1.push(root);
        que2.push(mirrorTree);
        while(!que1.empty() && !que2.empty()) {
            TreeNode *p1 = que1.front(), *p2 = que2.front();
            que1.pop(); que2.pop();
            if(p1->val != p2->val) return false;
            if(!pushChildNode(p1, p2, que1, que2)) return false;
        }
        if(!que1.empty() || !que2.empty()) return false;
        return true;

    }
};