103. Binary Tree Zigzag Level Order Traversal

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,null,null,15,7],

3 / \ 9 20 / \ 15 7

return its zigzag level order traversal as:

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

题意:

返回二叉树的“之”字形遍历结果

思路:

和普通的层次遍历思路类似，区别在于处理时需要维护两个栈，相邻两行分别存到两个栈中，进栈的顺序也不相同，一个栈是先进左子结点然后右子节点，另一个栈是先进右子节点然后左子结点，这样出栈的顺序就是我们想要的之字形了。记住后进栈的出站时先被处理：

/** * 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> >res; if (!root) return res; stack<TreeNode*> s1; stack<TreeNode*> s2; s1.push(root); vector<int> out; while (!s1.empty() || !s2.empty()) { while (!s1.empty()) { TreeNode *cur = s1.top(); s1.pop(); out.push_back(cur->val); if (cur->left) s2.push(cur->left); if (cur->right) s2.push(cur->right); } if (!out.empty()) res.push_back(out); out.clear(); while (!s2.empty()) { TreeNode *cur = s2.top(); s2.pop(); out.push_back(cur->val); if (cur->right) s1.push(cur->right); if (cur->left) s1.push(cur->left); } if (!out.empty()) res.push_back(out); out.clear(); } return res; } };