[LeetCode] Binary Tree Level Order Traversal

最新推荐文章于 2021-11-17 20:57:12 发布

1.02^365的成长裂变

最新推荐文章于 2021-11-17 20:57:12 发布

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分类专栏： LeetCode1

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本文提供了一种解决二叉树层次遍历问题的方法，并通过C++代码实现。该算法从根节点开始，按层级顺序从左到右返回每个节点的值。

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[Problem]

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

return its level order traversal as:

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

[Solution]

class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > res;
        if(NULL == root)return res;
   
        // insert root node
        vector<vector<TreeNode*> > nodeHeap;
        vector<int> rootLevVal;
        vector<TreeNode*> rootLevNode;
        rootLevVal.push_back(root->val);
        rootLevNode.push_back(root);
   
        res.push_back(rootLevVal);
        nodeHeap.push_back(rootLevNode);
   
        // level order
        int it = 0;
        int size = 1;
        while(it < size){
            vector<TreeNode*> tmp;
            tmp = nodeHeap[it];
   
            // level result
            vector<int> levelVal;
            vector<TreeNode*> levelNode;
   
            vector<TreeNode*>::iterator i;
            for(i = tmp.begin(); i != tmp.end(); ++i){
                if((*i)->left != NULL){
                    levelVal.push_back((*i)->left->val);
                    levelNode.push_back((*i)->left);
                }
                if((*i)->right != NULL){
                    levelVal.push_back((*i)->right->val);
                    levelNode.push_back((*i)->right);
                }
            }
            if(levelVal.size() > 0){
                res.push_back(levelVal);
                nodeHeap.push_back(levelNode);
                size++;
            }
            it++;
        }
        return res;
    }
};