LeetCode-----第103题-----二叉树的锯齿形层次遍历

二叉树的锯齿形层次遍历

难度:中等

给定一个二叉树,返回其节点值的锯齿形层次遍历。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。

例如:
给定二叉树 [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回锯齿形层次遍历如下:

[
  [3],
  [20,9],
  [15,7]
]

题目分析:

        使用两个队列,但是偶数队列存储的时候从左到右存储,奇数队列存储的时候从右到左

参考代码:

/**
 * 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<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int> > res;
        if(root == NULL)
            return res;
        
        stack<TreeNode*> odd_stack;
        stack<TreeNode*> even_stack;
        vector<int> temp;
        TreeNode* pMove = root;
        odd_stack.push(root);
        while(!odd_stack.empty() || !even_stack.empty())
        {
            while(!odd_stack.empty())
            {
                pMove = odd_stack.top();
                odd_stack.pop();
                temp.push_back(pMove->val);
                //偶数队列从左到右存储
                if(pMove->left)
                    even_stack.push(pMove->left);
                if(pMove->right)
                    even_stack.push(pMove->right);
            }
            if(!temp.empty())
            {
                res.push_back(temp);
                temp.clear();
            }
            while(!even_stack.empty())
            {
                pMove = even_stack.top();
                even_stack.pop();
                temp.push_back(pMove->val);
                //奇数队列从左到右存储
                if(pMove->right)
                    odd_stack.push(pMove->right);
                if(pMove->left)
                    odd_stack.push(pMove->left);
            }
            if(!temp.empty())
            {
                res.push_back(temp);
                temp.clear();
            }
        }
        return res; 
    }
};

猜你喜欢

转载自blog.csdn.net/L_smartworld/article/details/107407697
今日推荐