Find the sum of all left leaves in a given binary tree
每个节点判断左儿子是否是叶子, 是就把它值加上.
/**
* 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) {
int sum = 0;
if (root == null) {
return sum;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int length = queue.size();
for (int i = 0; i < length; i++) {
TreeNode node = queue.poll();
if (node.left != null) {
queue.add(node.left);
}
if (node.right != null) {
queue.add(node.right);
}
if (node.left != null && node.left.left == null && node.left.right == null) {
sum += node.left.val;
}
}
}
return sum;
}
}
时间复杂度O(n).