leetcode刷题，总结，记录，备忘 107

最新推荐文章于 2024-06-05 18:14:23 发布

guicaisa

最新推荐文章于 2024-06-05 18:14:23 发布

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本文介绍了解决LeetCode 107题的方法，即从叶节点到根节点的层次遍历二叉树。通过使用队列和栈容器来实现算法，记录每层节点数量，并按层次顺序收集节点值。

leetcode107Binary Tree Level Order Traversal II

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

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

return its bottom-up level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

不是很难的题目，请一定学会使用队列，栈容器，十分有用。通过一个count变量记录每层节点个数，并将一层的所有节点的值域放入数组之后，将一个数组放入二维数组中去，然后清空数组，继续同样的操作，最后将二维数组反序返回即可。

/**
 * 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>> levelOrderBottom(TreeNode* root) {
        if (!root) {
            return vector<vector<int> >();
        }
        vector<vector<int> > vi;
        queue<TreeNode*> st;
        st.push(root);
        int count = 1;
        vector<int> vti;
        while (!st.empty()) {
            TreeNode * tempTreeNode = st.front();
            st.pop();
            vti.push_back(tempTreeNode->val);
            if (tempTreeNode->left) {
                st.push(tempTreeNode->left);
            }
            if (tempTreeNode->right) {
                st.push(tempTreeNode->right);
            }
            count--;
            if (count == 0) {
                vi.push_back(vti);
                count = st.size();
                vti.clear();
             }
        }
        return vector<vector<int> >(vi.rbegin(), vi.rend());
    }
};