LeetCode –Binary Tree Preorder Traversal

题目:

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

Example:

Input:

 [1,null,2,3]
   1
    \
     2
    /
   3

Output:

 [1,3,2]

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

解题:

       遍历顺序“根-左-右”

实现:

  • 使用栈实现
         从根节点开始,先将根节点压入栈,保存结果值,然后移动到左节点, 保证“根-左”顺序, 为空后出栈,取其右节点入栈中。这样就保证了访问顺序为“根-左-右”。

/**
 * 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> res;
        stack<TreeNode*> ss;
        TreeNode *p = root;
        while (!ss.empty()||p){
            if (p!=NULL){
                ss.push(p);
                res.push_back(p->val);
                p = p->left;
            }
            else{
                p = ss.top();
                ss.pop();
                p=p->right;
                
            }
        }
        return res;
        
    }
};

实现:

  • 递归实现
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        
        vector<int> res;
        preorder(root, res);
        return res;
        
    }
    void preorder(TreeNode *root, vector<int> &res){
        if (!root) return;
        res.push_back(root->val);
        preorder(root->left, res);
        preorder(root->right, res);
    }
};

参考:

[LeetCode] 144. Binary Tree Preorder Traversal 二叉树的先序遍历

Tree Preorder Traversal 3 different solutions

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