#Leetcode# 103. Binary Tree Zigzag Level Order Traversal

发布时间：2020-12-14 04:26:59 所属栏目：大数据来源：网络整理

导读：https://leetcode.com/problems/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 betwe

https://leetcode.com/problems/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,15,7],

    3
   /   9  20
    /     15   7

return its zigzag level order traversal as:

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

代码：

/**
 * 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> > ans;
        if(!root) return ans;
        level(root,ans,1);
        for(int i = 0; i < ans.size(); i ++) {
            if(i % 2) {
                for(int j = 0; j < ans[i].size() / 2; j ++)
                    swap(ans[i][j],ans[i][ans[i].size() - j - 1]);
            }
        }
        return ans;
    }
    void level(TreeNode* root,vector<vector<int> >& ans,int depth) {
        vector<int> v;
        if(depth > ans.size()) {
            ans.push_back(v);
            v.clear();
        }
        ans[depth - 1].push_back(root -> val);
        if(root -> left)
            level(root -> left,depth + 1);
        if(root -> right)
            level(root -> right,depth + 1);
    }
};

（编辑：李大同）

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