404 Sum of Left Leaves 左叶子之和
Description:
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.
题目描述:
计算给定二叉树的所有左叶子之和。
示例：
3
/ \
9 20
/ \
15 7
在这个二叉树中，有两个左叶子，分别是 9 和 15，所以返回 24
思路:
- 递归, 找到结点的左子树非空且左子树的子树为空即为左叶子
- 迭代, 空间复杂度略优于递归方法
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
/**
* 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)
{
int result = 0;
if (!root) return 0;
vector<TreeNode*> nodes;
nodes.push_back(root);
while (nodes.size())
{
TreeNode* temp = nodes[0];
nodes.erase(nodes.begin());
if (temp -> left and !temp -> left -> left && !temp -> left -> right) result += temp -> left -> val;
if (temp -> left) nodes.push_back(temp -> left);
if (temp -> right) nodes.push_back(temp -> right);
}
return result;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int sumOfLeftLeaves(TreeNode root) {
return root == null ? 0 : (root.left != null && root.left.left == null && root.left.right == null) ? root.left.val + sumOfLeftLeaves(root.right) : sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right);
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def sumOfLeftLeaves(self, root: TreeNode) -> int:
return 0 if not root else root.left.val + self.sumOfLeftLeaves(root.right) if root.left and not root.left.left and not root.left.right else self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)
