lintcode:Binary Tree Preorder Traversal

原创

mb63887cf57331d 2022-12-01 19:05:28 ©著作权

©著作权归作者所有：来自51CTO博客作者mb63887cf57331d的原创作品，请联系作者获取转载授权，否则将追究法律责任

Given a binary tree, return the preorder traversal of its nodes’ values.
Can you do it without recursion?
http://www.lintcode.com/en/problem/binary-tree-preorder-traversal/

1.递归

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in vector which contains node values.
     */
    void Traversal(TreeNode *root,vector<int> &res){
        if(root==NULL){
            return;
        }
        else if(root){
            res.push_back(root->val);
        }
        if(root->left){
            Traversal(root->left,res);
        }
        if(root->right){
            Traversal(root->right,res);
        }
    }

    vector<int> preorderTraversal(TreeNode *root) {
        // write your code here
        vector<int> res;

        Traversal(root,res);

        return res;
    }
};

2.迭代
用迭代法做深度优先搜索的技巧就是使用一个显式声明的Stack存储遍历到节点，替代递归中的进程栈。对于先序遍历，我们pop出栈顶节点，记录它的值，然后将它的左右子节点push入栈，以此类推。

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in vector which contains node values.
     */
    vector<int> preorderTraversal(TreeNode *root) {
        // write your code here

        vector<int> res;

        stack<TreeNode*> s;

        if(root){
            s.push(root);
        }

        while(!s.empty()){
            TreeNode *node=s.top();
            s.pop();
            res.push_back(node->val);
            if(node->right){
               s.push(node->right);
            }
            if(node->left){
                s.push(node->left);
            }

        }

        return res;
    }
};