Binary Tree Zigzag Level Order Traversal

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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},

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

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

思路：层序遍历的改进。

/** * 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) {        deque<TreeNode*> Q1,Q2;        vector<vector<int> > res;        res.clear();        if (root == NULL) {            return res;        }        Q1.push_back(root);        vector<int> perRes;        perRes.push_back(Q1.back()->val);        res.push_back(perRes);        int i = 0;        while(!Q1.empty()) {            Q2.clear();            while(!Q1.empty()) {                TreeNode* pNode = Q1.front();                Q1.pop_front();                if (pNode->left != NULL) {                    Q2.push_back(pNode->left);                }                if (pNode->right != NULL) {                    Q2.push_back(pNode->right);                }            }            perRes.clear();            i++;            Q1 = Q2;            if (i&1) {                while(!Q2.empty()) {                    perRes.push_back(Q2.back()->val);                    Q2.pop_back();                }            }            else {                while(!Q2.empty()) {                    perRes.push_back(Q2.front()->val);                    Q2.pop_front();                }            }            if (!perRes.empty()) {                res.push_back(perRes);            }        }        return res;    }};

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