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.
Thoughts
检查当前结点的左子树是不是叶子.
Code
/**
* 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) {
if (root == null) {
return 0;
}
int sum = 0;
if (root.left != null && root.left.left == null && root.left.right == null) {
sum += root.left.val;
} else {
sum += sumOfLeftLeaves(root.left);
}
sum += sumOfLeftLeaves(root.right);
return sum;
}
}
Analysis
时间复杂度O(N).
Ver.2
遍历检查当前节点是否是左叶子.
/**
* 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) {
if (root == null) {
return 0;
}
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
int res = 0;
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
if (node.right != null) {
stack.push(node.right);
}
if (node.left != null) {
if (node.left.left == null && node.left.right == null) {
res += node.left.val;
} else {
stack.push(node.left);
}
}
}
return res;
}
}