LeetCode Binary Tree Traversal

1. 非递归的先序遍历

Given a binary tree, return the preorder traversal of its nodes' values.

For example:

Given binary tree

{1,#,2,3}

,

1
\
2
/
3

return

[1,2,3]

.

Note: Recursive solution is trivial, could you do it iteratively?

#include <vector>
#include <stack>

using namespace std;

struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> ivec;
stack<TreeNode*> treestk;
if(root == NULL)
return ivec;
TreeNode* curr = root;
while(curr != NULL || !treestk.empty()) {
while(curr != NULL) {
ivec.push_back(curr->val);
treestk.push(curr);
curr = curr->left;
}
curr = (treestk.top())->right;
treestk.pop();
}
return ivec;
}
};

2. 非递归的中序遍历

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode *root) {
vector<int> ivec;
if(root == NULL)
return ivec;
stack<TreeNode*> treestk;
TreeNode* curr = root;
while(curr != NULL || !treestk.empty()) {
while(curr != NULL) {
treestk.push(curr);
curr = curr->left;
}
ivec.push_back((treestk.top())->val);
curr = (treestk.top())->right;
treestk.pop();
}
}
};

3. 非递归的后序遍历

#include <vector>
#include <stack>

using namespace std;

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) {
vector<int> ivec;
stack<pair<TreeNode*, bool>> treestk;
if(root == NULL)
return ivec;
treestk.push(make_pair(root, false));
while(!treestk.empty()) {
auto& curr = treestk.top();
if(curr.second) {
ivec.push_back((curr.first)->val);
treestk.pop();
}
else {
curr.second = true;
if((curr.first)->right != NULL) {
treestk.push(make_pair((curr.first)->right,false));
}
if((curr.first)->left != NULL) {
treestk.push(make_pair((curr.first)->left,false));
}
}
}
}
};