Binary Tree Leaf Sum
ID: 481; easy; 二叉树叶子节点之和
Solution 1 (Java)
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: the root of the binary tree
* @return: An integer
*/
public int leafSum(TreeNode root) {
if (root == null)
return 0;
if (root.left == null && root.right == null)
return root.val;
return leafSum(root.left) + leafSum(root.right);
}
}
Notes
Divide and conquer
Solution 2 (Java)
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
private int sum;
/**
* @param root: the root of the binary tree
* @return: An integer
*/
public int leafSum(TreeNode root) {
sum = 0;
leafSumHelper(root);
return sum;
}
private void leafSumHelper(TreeNode root) {
if (root == null) return;
if (root.left == null && root.right == null) {
sum += root.val;
return;
}
leafSumHelper(root.left);
leafSumHelper(root.right);
}
}
Notes
Traversal
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