[LeetCode]103. Binary Tree Zigzag Level Order Traversal--二叉树之字形遍历

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]
]

---

分析：简单分析就可以发现，每行都是先进后出，所以很显然需要用到栈。但需要注意一点，对于每一行，左右孩子入栈的顺序不同，需要专门标记一下，奇数行先左孩子后右孩子入栈，偶数行先右孩子后左孩子入栈。

/** * 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>> result;        if(root == NULL)            return result;        stack<TreeNode*>current, next;        current.push(root);        vector<int> oneLayer;        bool flag = true;        while(!current.empty()){            TreeNode* pCur = current.top();            current.pop();            oneLayer.push_back(pCur->val);            // 需要确定左右子树入栈的顺序            // 用flag标记            if(flag){                if(pCur->left)                    next.push(pCur->left);                if(pCur->right)                    next.push(pCur->right);            }else{                if(pCur->right)                    next.push(pCur->right);                if(pCur->left)                    next.push(pCur->left);            }            if(current.empty()){                result.push_back(oneLayer);                oneLayer.clear();            // 清空                swap(current, next);         // 直接交换                flag = !flag;                // flag切换            }        }        return result;    }};

也可以使用《剑指offer》中解法，思想是一样的，代码略微不同。

/** * 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>> result;        if(!root)            return result;        stack<TreeNode*> s[2];        int current = 0;        int next = 1;        s[current].push(root);        vector<int> v;        while(!s[current].empty() || !s[next].empty()){            TreeNode* pCur = s[current].top();            s[current].pop();            v.push_back(pCur->val);            if(current == 0){                if(pCur->left)                    s[next].push(pCur->left);                if(pCur->right)                    s[next].push(pCur->right);            }else{                if(pCur->right)                    s[next].push(pCur->right);                if(pCur->left)                    s[next].push(pCur->left);            }            if(s[current].empty()){                current = 1 - current;                next = 1 - next;                result.push_back(v);                v.clear();            }        }        return result;    }};