标签：des style blog io strong for ar div cti
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);
}
};
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;
}
};
38. Same Tree && Symmetric Tree
标签：des style blog io strong for ar div cti
原文地址：http://www.cnblogs.com/liyangguang1988/p/3940137.html
