LeetCode 404. Sum of Left Leaves(二叉树)
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Find the sum of all left leaves in a given binary tree.
Example:
3 / \ 9 20 / \ 15 7There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
思路:
二叉树问题最简单的方法是使用递归,关于递归算法的设计要注意两点:1)确定递归公式;2)确定边界条件(即递归出口)。本题目中,递归出口为:当左孩子节点为叶子节点时,将其左孩子节点的值加入sum中;或者若节点不存在,将0加入sum中。判断一个节点是否为叶子节点,条件为该节点非空,且其左孩子和右孩子节点为空。
所以该题的思路为:判断节点的左孩子节点是否为叶子节点,若左孩子节点为叶子节点,则返回左孩子节点的值+右子树的递归结果;若左孩子节点为非叶子节点,则返回左子树的递归结果+右子树的递归结果。
Code:
/** * 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); }};
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