題目:
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,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]題目鏈接
題意:
給一個二叉樹,要求按照從底層到頂層的順序遍歷整個二叉樹,每一層的節點從左到右輸出。
在遞歸的同時維護一個變量depth,表示當前的深度,每次向表示當前深度的vector添加節點,最後把ans反轉一下順序,就變成了從底層項頂層的順序。
代碼如下:
class Solution {
public:
vector<vector<int>> ans;
void dfs(TreeNode* node, int depth) {
if (!node) return;
if (ans.size() <= depth) {
ans.push_back({node->val});
}
else {
ans[depth].push_back(node->val);
}
dfs(node->left, depth+1);
dfs(node->right, depth+1);
}
vector<vector<int>> levelOrderBottom(TreeNode* root) {
dfs(root, 0);
reverse(ans.begin(), ans.end());
return ans;
}
};