Given a binary tree, return all root-to-leaf paths.
For example, given the following binary tree:
1
/ \
2 3
\
5
All root-to-leaf paths are:
["1->2->5", "1->3"]
思路：用两个stack<TreeNode*> in , s;
in : 记录当前的路径 p , 和vector<int>path 相同，只不过一个记录的是 val ，一个记录点的指针。
s : 记录每个节点的 p->right
v2s : 将path转换称符合要求的string
/**
* 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:string v2s(vector<int>a){
const int len = a.size();
string re = "";
char *p = new char[];
for (int i = ; i<len; i++){
stringstream ss; // 需要包含头文件 #include<sstream>
string temp;
ss << a[i];
ss >> temp;
re = re + temp;
if (i != len - )
re = re + "->";
}
return re;
}
vector<string> binaryTreePaths(TreeNode* root) {
vector<int> path;
vector<string> re;
if (root == NULL) return re;
TreeNode * p = root;
stack<TreeNode*> in, s;
while (p != NULL){
if (p->left == NULL&&p->right == NULL){
in.push(p);
path.push_back(p->val);
re.push_back(v2s(path));
if (s.empty())
return re;
else {
// 通过while循环，查找下一个需要遍历的点
while (!in.empty() && !s.empty() && (in.top())->right != s.top()){
in.pop();
path.erase(path.end()-);
}
p = s.top();
s.pop();
}
}
else if (p->left == NULL&&p->right != NULL){
path.push_back(p->val);
in.push(p);
p = p->right;
}
else if (p->left != NULL&&p->right == NULL){
path.push_back(p->val);
in.push(p);
p = p->left;
}
else{
path.push_back(p->val);
in.push(p);
s.push(p->right);
p = p->left;
}
}
return re;
}
};
另一道相似题目
有一棵二叉树，树上每个点标有权值，权值各不相同，请设计一个算法算出权值最大的叶节点到权值最小的叶节点的距离。二叉树每条边的距离为1，一个节点经过多少条边到达另一个节点为这两个节点之间的距离。
给定二叉树的根节点root，请返回所求距离。
思路：只需要将保存的路径进行比较，相同的去掉然后相加，就能算出路径长度。
/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};*/
class Tree {
public:
int getDis(TreeNode* root) {
// write code here
stack<TreeNode*> max, min;
int m = INT_MIN, n = INT_MAX, re = 0;
stack<TreeNode*> in, s;
TreeNode *p = root;
while (p != NULL){
if (p->left == NULL&&p->right == NULL){
in.push(p);
if (p->val > m){
max = in;
m = p->val;
}
if (p->val < n){
min = in;
n = p->val;
}
while (!in.empty() && !s.empty() && (in.top())->right != s.top()){
in.pop();
}
if (s.empty())
break;
else {
p = s.top();
s.pop();
}
}
else if (p->left == NULL&&p->right != NULL){
in.push(p);
p = p->right;
}
else if (p->left != NULL&&p->right == NULL){
in.push(p);
p = p->left;
}
else {
in.push(p);
s.push(p->right);
p = p->left;
}
}
stack<TreeNode*> t1, t2;
while (!max.empty()){
t1.push(max.top());
max.pop();
}
while (!min.empty()){
t2.push(min.top());
min.pop();
}
while (!t1.empty() && !t2.empty()){
if (t1.top() != t2.top())
break;
else
t1.pop(); t2.pop();
}
re = re + t1.size() + t2.size();
return re;
}
};