[leetcode-144-Binary Tree Preorder Traversal]

Posted 2020-09-01 hellowOOOrld

Given a binary tree, return the preorder traversal of its nodes‘ values.
Given binary tree {1,#,2,3},
　 1
　　\
　　 2
　　/
　 3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?

如下是非递归版本。

    void preorderNonRecursiveTraversal(TreeNode*pRoot, vector<int>& result)
    {
        if (pRoot == NULL) return;
        stack<TreeNode*> st;
        TreeNode* temp;
        st.push(pRoot);
        while (!st.empty())
        {
            temp = st.top();
            result.push_back(temp->val);
            st.pop();
            if (temp->right != NULL)st.push(temp->right);//先右 后左
            if (temp->left != NULL)st.push(temp->left);
        }

    }
    vector<int> preorderTraversal(TreeNode* root)
    {
        vector<int> result;
        if (root == NULL) return result;
        preorderNonRecursiveTraversal(root, result);
        return result;
    }

 递归版本：

void PreOrder(TreeNode* pRoot, vector<TreeNode*>& result)
    {
        if (pRoot != NULL)
        {
            result.push_back(pRoot);
            PreOrder(pRoot->left, result);            
            PreOrder(pRoot->right, result);
        }
    }