Lintcode481-Binary Tree Leaf Sum-Easy
481. Binary Tree Leaf Sum
Given a binary tree, calculate the sum of leaves.
Example
Example 1:
Input:
1
/ \
2 3
/
4
output:7.
Example 2:
Input:
1
\
3
Output:3.
注意:
sum是类成员变量的原因:因为没执行一次递归,递归方法里的局部变量都会被重新创建。在递归函数中,不需要每次都创建sum,只有访问叶子结点时,对sum操作即可。
递归过程详解(讲解很透彻,包括递归中的return过程,递归的终止条件)
理解递归,关键是脑中有一幅代码的图片,函数执行到递归函数入口时,就扩充一段完全一样的代码,执行完扩充的代码并return后,继续执行前一次递归函数中递归函数入口后面的代码
递归法代码:
/**
* 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
*/
int sum = 0;
public int leafSum(TreeNode root) {
helper(root);
return sum;
}
public void helper(TreeNode root) {
if (root == null) {
return;
} if (root.left == null && root.right == null) {
sum += root.val;
return;
} helper(root.left);
helper(root.right);
}
}
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