二叉树-专题

题型一：非递归遍历二叉树后续

/**
* 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:
vector<int> postorderTraversal(TreeNode* root) {
map<TreeNode*,bool> mymap;
vector<int> ret;
stack<TreeNode*> mystac;
TreeNode* p=root;
while(p||!mystac.empty())
{
while(p)
{

mystac.push(p);
mymap[p]=false;
p=p->left;

}

if(!mystac.empty())
{
TreeNode* tmp=mystac.top();
mystac.pop();
if(mymap[tmp]==false)
{
mymap[tmp]=true;
mystac.push(tmp);
p=tmp->right;
}
else
{
ret.push_back(tmp->val);
p=NULL;
}
}
}

return ret;

}
};

　　题型二：非递归二叉序前序遍历（中序差不多，就不写了，自己去脑补去。。。。中序的逆序是直接先遍历右边再遍历左边，orz

vector<int> preorderTraversal(TreeNode *root) {
        // write your code here
        vector<int> ret;
        stack<TreeNode*> sta;
        TreeNode* p=root;
        while(sta.size()>0||p)
        {
            while(p)
            {
                sta.push(p);
                ret.push_back(p->val);
                p=p->left;
            }
            if(sta.size())
            {
                TreeNode* tmp=sta.top();
                sta.pop();
                p=tmp->right;
            }
        }
        return ret;
    }

　　二叉树中和为某一值的路径：

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    void dfs(TreeNode* root,int e,int sum,vector<vector<int>>& ret,vector<int> tmp)
    {

        if(root)
        {
            if(sum+root->val==e&&root->left==NULL&&root->right==NULL)
            {
                tmp.push_back(root->val);
                ret.push_back(tmp);
                return;
            }
            tmp.push_back(root->val);
            dfs(root->left,e,root->val+sum,ret,tmp);
            dfs(root->right,e,root->val+sum,ret,tmp);
        }

    }
    vector<vector<int> > FindPath(TreeNode* root,int expectNumber) {
        vector<vector<int>> ret;
        vector<int> tmp;
        dfs(root,expectNumber,0,ret,tmp);
        return ret;
    }
};