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sum-of-left-leaves

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Sum of Left Leaves – LeetCode 404

Problem

Description

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

Answer

Original

Code

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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) return 0;
        if(!root->left) return sumOfLeftLeaves(root->right);
        if(!root->right){
            if(root->left->left == root->left->right)
                return root->left->val;
            else 
                return sumOfLeftLeaves(root->left);
        }
        if(root->left->left == root->left->right)
                return root->left->val + sumOfLeftLeaves(root->right);
            else 
                return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
    }
};

思路

简单的用DFS查找左子叶而后返回加和。用BFS并不占优。时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$4$ ms，排名$58.70\%$

Better

思路

重写了代码，减少了不必要的分支。时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$4$ ms，排名$58.70\%$

Code

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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) return 0;
        if (root->left && !root->left->left && !root->left->right) {
            return root->left->val + sumOfLeftLeaves(root->right);
        }
        return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
    }
};