Binary Tree Preorder Traversal

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

Example:

Input: [1,null,2,3]
   1
    
     2
    /
   3

Output: [1,2,3]

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


typedef char ElemType; typedef struct BiTreeNode { ElemType data; struct BiTreeNode *left; struct BiTreeNode *right; }BiTreeNode,*BiTree; Binary Tree Preorder Traversal void TraverseBiTree(BiTree T){ if(T == NULL) return; printf("%c",T->data); TraverseBiTree(T->left); TraverseBiTree(T->right); }

//非递归方法待完善,以下方法便于理解,如果还是不太理解,不妨画图来看一下,见得多了,抽象能力自然就有了。enjoy!
Binary Tree Preorder Traversal
这个方法便于记忆。
class Solution{
    public:
        vector<int> preorderTraversal(TreeNode* root){
            if(!root) return {};
            vector<int> res;
            stack<TreeNode*> s{{root}};
            while(!s.empty()){
                TreeNode* t = s.top();s.pop();
                res.push_back(t->val);
                if(t->right) s.push(t->right);
                if(t->left) s.push(t->left);
            }
            return res;
        }
};

//这个方法可以拓展思路 c
++ solution2: class Solution{ public: vector<int> preorderTraversal(TreeNode* root){ vector<int> res; stack<TreeNode*> s; TreeNode *p = root; while(!s.empty() || p){ if(p){ s.push(p); res.push_back(p->val); p = p->left; } else{ TreeNode *t = s.top();s.pop(); p = t->right; } } return res; } };


 
怕什么真理无穷,进一寸有一寸的欢喜。---胡适
原文地址:https://www.cnblogs.com/hujianglang/p/11421184.html