算法-数据结构之二叉树

1.翻转二叉树

在翻转问题中总结了

地址：http://www.cnblogs.com/vhyz/p/7241743.html

2.判断二叉树是否相等

2.1 100. 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.

2.2算法初步

利用递归的思路，若两者不同时为空，则返回false，若同时为空，则返回true

因为如果进行到了最后一步都为空，那么一定为true

/**
 * 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 isSameTree(TreeNode* p, TreeNode* q) {
        if(((p==nullptr)^(q==nullptr)))
           return false;
        if(p==nullptr&&q==nullptr)
            return true;
        if(p->val==q->val)
           return (isSameTree(p->left,q->left)&&isSameTree(p->right,q->right));
            else return false;
    }
};

2.3优化算法

注意到前两个判断条件可以合并

/**
 * 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 isSameTree(TreeNode* p, TreeNode* q) {
        if(p==nullptr||q==nullptr)
            return p==q;
           return (p->val==q->val&&isSameTree(p->left,q->left)&&isSameTree(p->right,q->right));
    }
};

3.判断是否为镜面二叉树

3.1 题目 Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/
2 2
/ \ /
3 4 4 3

But the following [1,2,2,null,3,null,3] is not:
1
/
2 2
\
3 3

3.2 queue算法

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(root==nullptr)
            return true;
        queue<TreeNode*>s1,s2;
        TreeNode*left,*right;
        s1.push(root->left);
        s2.push(root->right);
        while(!s1.empty()&&!s2.empty())
        {
            left=s1.front();
            s1.pop();
            right=s2.front();
            s2.pop();
            if(left==nullptr&&right==nullptr)
                continue;
            if(left==nullptr||right==nullptr)
                return false;
            if(left->val!=right->val)
                return false;
            s1.push(left->left);
            s1.push(left->right);
            s2.push(right->right);
            s2.push(right->left);
        }
        return true;
    }
};