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
     
       & qu, queue
      
       & 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
       
         que1;        queue
        
          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;            }};