[LeetCode] 404. Sum of Left Leaves

轉自LeetCode

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.

<Solution>

通常 tree 的問題，不是 DFS 就是 BFS

那實作的方式，就有 iterative 和 recursive 兩種

這邊就按照提意來寫就可以

code 如下

C++，Iterative

	/**
	* 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) {
	stack<TreeNode *> stack;
	if(root) {
	stack.push(root);
	}
	int ans = 0;
	TreeNode *cur;
	while(!stack.empty()) {
	cur = stack.top();
	stack.pop();
	if(cur->left) {
	if(!cur->left->left && !cur->left->right) {
	//>> left leaf
	ans += cur->left->val;
	}
	else {
	//>> left childs
	stack.push(cur->left);
	}
	}
	//>> only push right child
	if(cur->right && (cur->right->left || cur->right->right)) {
	stack.push(cur->right);
	}
	}
	return ans;
	}
	};

view raw sumOfLeftLeaves_1.cpp hosted with ❤ by GitHub

C++，Recursive

	/**
	* 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) {
	return findSum(root, false);
	}
	int findSum(TreeNode *node, bool isLeft) {
	if(!node) {
	return 0;
	}
	if(!node->left && !node->right) {
	return isLeft ? node->val : 0;
	}
	return findSum(node->left, true) + findSum(node->right, false);
	}
	};

view raw sumOfLeftLeaves_2.cpp hosted with ❤ by GitHub