A404-Sum-of-Left-Leaves

题目描述

Find the sum of all left leaves in a given binary tree.

Example:

3

/ \
9 20
/ \
15 7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

我的解法

解题思路

求二叉树的左子叶的和,可以通过递归的方法(深度优先):

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/**
* 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 {
private:
int res;
public:
int sumOfLeftLeaves(TreeNode* root) {
if (root == NULL) return 0;
if (root -> left && root -> left -> left == NULL && root -> left -> right == NULL)
res += root -> left -> val;
if (root -> left) sumOfLeftLeaves(root -> left);
if (root -> right) sumOfLeftLeaves(root -> right);
return res;
}
};

执行用时:8 ms, 在所有 C++ 提交中击败了51.83%的用户
内存消耗:13.1 MB, 在所有 C++ 提交中击败了71.76%的用户

再试试迭代的方法(广度优先):

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/**
* 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:
int sumOfLeftLeaves(TreeNode* root) {
if (root == NULL) return 0;
queue<TreeNode*> q;
q.push(root);
int res = 0;
while(!q.empty()){
int n = q.size();
for (int i = 0; i < n; i++){
TreeNode* node = q.front();
q.pop();
if (node -> left && node -> left -> left == NULL && node -> left -> right == NULL){
res += node -> left -> val;
}
if (node -> left) q.push(node -> left);
if (node -> right)q.push(node ->right);
}
}
return res;
}
};

执行用时:4 ms, 在所有 C++ 提交中击败了89.39%的用户
内存消耗:13.3 MB, 在所有 C++ 提交中击败了14.86%的用户