144 Binary Tree Preorder Traversal（二叉樹的前序走訪）+（二叉樹、迭代）

標籤：

翻譯

給定一個二叉樹，返回其前序走訪的節點的值。例如：給定二叉樹為 {1，#， 2， 3}   1         2    /   3返回 [1, 2, 3]備忘：用遞迴是微不足道的，你可以用迭代來完成它嗎？

原文

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

分析

題目讓咱試試迭代呢，不過還是先老老實實把遞迴給寫出來再說吧~

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<int> v;    vector<int> preorderTraversal(TreeNode* root) {        if (root != NULL) {            v.push_back(root->val);            preorderTraversal(root->left);            preorderTraversal(root->right);        }        return v;    }};

緊接著，咱來寫迭代的……

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/class Solution {public:    vector<int> preorderTraversal(TreeNode *root) {        vector<int> result;        stack<TreeNode*> tempStack;        while (!tempStack.empty() || root != NULL) {            if (root != NULL) {                result.push_back(root->val);                tempStack.push(root);                root = root->left;            }            else {                root = tempStack.top();                tempStack.pop();                root = root->right;            }        }        return result;    }};

144 Binary Tree Preorder Traversal（二叉樹的前序走訪）+（二叉樹、迭代）