LeetCode 404. Sum of Left Leaves

标签：ftl   binary   leetcode   value   spec   递归   and   null   tree   

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.

/**
 * 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) { // 树的递归遍历
        if(root==NULL) return 0;
        if(root->left!=NULL&&root->left->left==NULL&&root->lef;
        return sum;t->right==NULL)
           sum+=root->left->val;
        //if(root->left!=NULL)
           sumOfLeftLeaves(root->left);
        //if(root->right!=NULL)
           sumOfLeftLeaves(root->right)
    }
private:
    int sum=0;
};

LeetCode 404. Sum of Left Leaves

标签：ftl   binary   leetcode   value   spec   递归   and   null   tree   

原文地址：https://www.cnblogs.com/A-Little-Nut/p/10066849.html