LeetCode 103. Binary Tree Zigzag Level Order Traversal

最新推荐文章于 2024-01-19 11:15:23 发布

原创最新推荐文章于 2024-01-19 11:15:23 发布 · 167 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 题解专栏收录该内容

288 篇文章

订阅专栏

本文介绍了一种使用栈实现的二叉树锯齿形层序遍历算法，通过控制方向标志实现从左到右再从右到左的交替遍历方式，并给出了具体的C++代码实现。

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]
]

Just use a stack and a flag..... But it is very easy to make mistakes....

    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        if(!root) return {};
        stack<TreeNode*> nodes;
        nodes.push(root);
        vector<vector<int>> res;
        bool leftToRight = true;
        while(!nodes.empty()) {
            vector<int> level;
            stack<TreeNode*> nextLevel;
            while(!nodes.empty()) {
                TreeNode* tmp = nodes.top();
                nodes.pop();
                level.push_back(tmp->val);
                if(leftToRight) {
                    if(tmp->left)
                    nextLevel.push(tmp->left);
                    if(tmp->right)
                    nextLevel.push(tmp->right);
                } else {
                    if(tmp->right)
                    nextLevel.push(tmp->right);
                    if(tmp->left)
                    nextLevel.push(tmp->left);
                }
            }
            leftToRight = !leftToRight;
            res.push_back(level);
            nodes.swap(nextLevel);
        }
        return res;