【LeetCode】144. 二叉树的前序遍历-递归和非递归

LeetCode链接:https://leetcode-cn.com/problems/binary-tree-preorder-traversal/

给定一个二叉树,返回它的 前序 遍历。

 示例:

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

输出: [1,2,3]


一,非递归解决方案:

     根据前序遍历访问的顺序,优先访问根结点,然后再分别访问左孩子和右孩子。即对于任一结点,其可看做是根结点,因此可以直接访问,访问完之后,若其左孩子不为空,按相同规则访问它的左子树;当访问其左子树时,再访问它的右子树。因此其处理过程如下:

  对于任一结点cur:

     1)访问结点cur,并将结点cur入栈;

     2)判断结点cur的左孩子是否为空,若为空,则取栈顶结点并进行出栈操作,并将栈顶结点的右孩子置为当前的结点cur,循环至1);若不为空,则将cur的左孩子置为当前的结点cur;

     3)直到cur为NULL并且栈为空,则遍历结束。

/**
 * 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:
    /*void preorder(TreeNode* root,vector& v)
    {
        if(root == 0)
            return;
        v.push_back(root->val);
        preorder(root->left,v);
        preorder(root->right,v);
    }*/
    vector preorderTraversal(TreeNode* root) 
    {
        vector v;
        //preorder(root,v);
        TreeNode* cur = root;
        stack st;
        while(cur || !st.empty())
        {
            while(cur)
            {
                st.push(cur);
                v.push_back(cur->val);
                cur = cur->left;
            }
            TreeNode* top = st.top();
            st.pop();
            cur = top->right;
        }
        return v;
    }
};

  【LeetCode】144. 二叉树的前序遍历-递归和非递归_第1张图片

二,进阶: 递归算法很简单,你可以通过迭代算法完成吗?

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int *p;
int size;
void _pre(struct TreeNode *root){
    if(root == NULL){
        return;
    }
    p[size] = root->val;
    size++;
    _pre(root->left);
    _pre(root->right);
}
int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    p = (int *)malloc(sizeof(int)*100000);
    size = 0;
    _pre(root);
    *returnSize = size;
    return p;
}

三,Python方案:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root == None:
            return[]
        return
[root.val]+self.preorderTraversal(root.left)+self.preorderTraversal(root.right)

你可能感兴趣的:(LeetCode)