Leetcode: Binary Tree Preorder Traversal

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?

Recursive:

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/**

* Definition for binary tree

* struct TreeNode {

* int val;

* TreeNode *left;

* TreeNode *right;

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

* };

*/

class Solution {

public:

void preorder(TreeNode* root, vector<int> &path)

{

if(root!=NULL)

{

path.push_back(root->val);

preorder(root->left, path);

preorder(root->right, path);

}

}

vector<int> preorderTraversal(TreeNode *root) {

// IMPORTANT: Please reset any member data you declared, as

// the same Solution instance will be reused for each test case.

vector<int> path;

preorder(root, path);

return path;

}

};

Iterative:

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vector<int> preorderTraversal(TreeNode *root) {

// IMPORTANT: Please reset any member data you declared, as

// the same Solution instance will be reused for each test case.

vector<int> path;

stack<TreeNode*> stk;

while(root != NULL || !stk.empty())

{

if(root != NULL)

{

while(root)

{

path.push_back(root->val);

stk.push(root);

root=root->left;

}

}

else{

root = stk.top()->right;

stk.pop();

}

}

return path;

}

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vector<int> preorderTraversal(TreeNode *root) {

// IMPORTANT: Please reset any member data you declared, as

// the same Solution instance will be reused for each test case.

vector<int> path;

stack<TreeNode*> stk;

if(root != NULL)

{

stk.push(root);

while(!stk.empty())

{

root = stk.top();

stk.pop();

path.push_back(root->val);

if(root->right) stk.push(root->right);

if(root->left) stk.push(root->left);

}

}

return path;

}

其他两种非递归的遍历算法：

中序遍历-非递归：

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vector<int> InOrderTraversal(TreeNode *root) {

// IMPORTANT: Please reset any member data you declared, as

// the same Solution instance will be reused for each test case.

vector<int> path;

stack<TreeNode*> stk;

while(root != NULL || !stk.empty())

{

while(root != NULL)

{

stk.push(root);

root = root->left;

}

if( !stk.empty())

{

root = stk.top();

stk.pop();

root = root->right;

}

}



return path;

}

后序遍历-非递归：

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struct TNode{

//TreeNode* root;

int val;

TNode *left;

TNode *right;

bool isVisited;

TNode(int a):val(a),left(NULL),right(NULL),isVisited(false){};

};



vector<int> PostOrderTraversal(TNode *root) {

// IMPORTANT: Please reset any member data you declared, as

// the same Solution instance will be reused for each test case.

vector<int> path;

stack<TNode*> stk;

if(root != NULL)

{

root->isVisited = false;

stk.push(root);

}

TNode* cur = NULL;

while(!stk.empty())

{

cur = stk.top();

if(cur->isVisited)

{

path.push_back(cur->val);

stk.pop();

}else{

cur->isVisited=true;

if(cur->right)

{

cur->right->isVisited = false;

stk.push(cur->right);

}

if(cur->left)

{

cur->left->isVisited = false;

stk.push(cur->left);

}

}

}

return path;

}