LeetCode ---Binary Tree Level Order Traversal

最新推荐文章于 2024-06-25 11:36:41 发布

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本文介绍了一种解决二叉树层次遍历问题的方法。通过使用队列存储节点及其层级信息，实现从左到右逐层遍历节点，并返回各层节点值。示例展示了如何构造二叉树并获取层次遍历结果。

Binary Tree Level Order Traversal

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},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

My Submitted Code

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

struct TreeNodeWithL{
    TreeNode* node;
    int level;
    TreeNodeWithL(TreeNode* n,int l):node(n),level(l){}
};
class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int> > result;
        vector<int> le;
        queue<TreeNodeWithL> q; //层次遍历使用队列
        if(!root){
            return result;
        }
        TreeNodeWithL node(root,0);
        q.push(node);
        int currlev=-1;
        while(!q.empty()){
            TreeNodeWithL node=q.front(); //每次取队列头，然后弹出
            q.pop();
            if(node.node->left){
                TreeNodeWithL lnode(node.node->left,node.level+1);
                q.push(lnode);
            }
            if(node.node->right){
                TreeNodeWithL rnode(node.node->right,node.level+1);
                q.push(rnode);
            }

            if(currlev!=node.level){
                if(currlev!=-1){
                    result.push_back(le);
                }
                currlev=node.level;
                le.clear();
                le.push_back(node.node->val);

            }else{
                le.push_back(node.node->val);
            }

        }
        result.push_back(le);
        return result;


    }
};