[LeetCode] Binary Tree Zigzag Level Order Traversal

最新推荐文章于 2021-02-05 11:10:39 发布

原创最新推荐文章于 2021-02-05 11:10:39 发布 · 548 阅读

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本文详细介绍了二叉树锯齿形层次遍历算法的两种递归解决方案和一种迭代解决方案，包括算法原理、代码实现及示例应用。

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree {3,9,20,#,#,15,7},

return its zigzag level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

recursive solution 1

/**
 * Definition for binary tree
 * 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>> result;
        if(!root)   return result;
        traverse(root, 1, result);
        int n = result.size();
        for(int i = 1; i < n; i += 2)
            reverse(result[i].begin(), result[i].end());
        return result;
    }
    void traverse(TreeNode *root, int level, vector<vector<int>> &result)
    {
        if(!root)   return;
        if(level > result.size())
            result.push_back(vector<int>());
        result[level-1].push_back(root->val);
        traverse(root->left, level+1, result);
        traverse(root->right, level+1, result);
    }
};

recursive solution 2

/**
 * Definition for binary tree
 * 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>> result;
        if(!root)   return result;
        traverse(root, 1, result, true);
        return result;
    }
    void traverse(TreeNode *root, int level, vector<vector<int>> &result, bool left2right)
    {
        if(!root)   return;
        if(level > result.size())
            result.push_back(vector<int>());
        if(left2right)
            result[level-1].push_back(root->val);
        else
            result[level-1].insert(result[level-1].begin(), root->val);
        traverse(root->left, level+1, result, !left2right);
        traverse(root->right, level+1, result, !left2right);
    }
};

iterative solution

/**
 * Definition for binary tree
 * 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>> result;
        if(!root)   return result;
        
        bool rev = false;
        queue<TreeNode*> current, next;
        vector<int> level;
        current.push(root);
        while(!current.empty())
        {
            while(!current.empty())
            {
                TreeNode *node = current.front();
                current.pop();
                level.push_back(node->val);
                if(node->left)  next.push(node->left);
                if(node->right) next.push(node->right);
            }
            if(rev)
                reverse(level.begin(), level.end());
            result.push_back(level);
            level.clear();
            rev = !rev;
            swap(next, current);
        }
        return result;
    }
};