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  • [LeetCode] Binary Tree Inorder Traversal Solution


    Given a binary tree, return the inorder traversal of its nodes' values.
    For example:
    Given binary tree {1,#,2,3},
       1
    \
    2
    /
    3
    return [1,3,2].
    Note: Recursive solution is trivial, could you do it iteratively?
    confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
    » Solve this problem

    [Thoughts]
    For recursion version, it's very easy to write.

    But for iterative version, we need a stack to help.


    [Code]
    Recursion version
    1:    vector<int> inorderTraversal(TreeNode *root) {  
    2: // Start typing your C/C++ solution below
    3: // DO NOT write int main() function
    4: vector<int> result;
    5: inorderTra(root, result);
    6: return result;
    7: }
    8: void inorderTra(TreeNode* node, vector<int> &result)
    9: {
    10: if(node == NULL)
    11: {
    12: return;
    13: }
    14: inorderTra(node->left, result);
    15: result.push_back(node->val);
    16: inorderTra(node->right, result);
    17: }

    Iteration version
    1:    vector<int> inorderTraversal(TreeNode *root) {  
    2: // Start typing your C/C++ solution below
    3: // DO NOT write int main() function
    4: vector<TreeNode*> sta;
    5: vector<int> result;
    6: if(root == NULL) return result;
    7: TreeNode* node =root;
    8: while(sta.size()>0 || node!=NULL)
    9: {
    10: while(node!=NULL)
    11: {
    12: sta.push_back(node);
    13: node = node->left;
    14: }
    15: node= sta.back();
    16: sta.pop_back();
    17: result.push_back(node->val);
    18: node =node->right;
    19: }
    20: return result;
    21: }



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  • 原文地址:https://www.cnblogs.com/codingtmd/p/5078926.html
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