LeetCode #404 - Sum of Left Leaves - Easy
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Problem
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.
Algorithm
整理一下题意:给定一棵二叉树,求最左边节点之和。
考虑三种情况。
- 若根节点为NULL,则返回0
- 若左子节点为空但是右子不为空,则求和转移到右子数
- 若左子和左子均非空,则返回左子和右子的递归调用
于是可以得到递归方法的解。
代码如下。
//递归版本,用时3msclass 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); }};
同样的思路,也可以非递归地实现。通过层次遍历,将所有层次最左节点加入到和中。注意while中的三个if覆盖了递归版本的三种情况。
代码如下。
//非递归版本,用时6msclass Solution {public: int sumOfLeftLeaves(TreeNode* root) { if(root==NULL||root->left==NULL&&root->right==NULL) return 0; TreeNode* p; int sum=0; queue<TreeNode*> q; queue<TreeNode*> level; q.push(root); while(!q.empty()){ p=q.front(); q.pop(); if(p->left!=NULL&&p->left->left==NULL&&p->left->right==NULL) sum+=p->left->val; if(p->left!=NULL) q.push(p->left); if(p->right!=NULL) q.push(p->right); } return sum; }};
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