404. 左葉子之和
404. Sum of Left Leaves

LeetCode404. Sum of Left Leaves

題目描述
計算給定二叉樹的所有左葉子之和。

示例:

3/ \9 20/ \15 7

在這個二叉樹中，有兩個左葉子，分別是 9 和 15，所以返回 24。

Java 實現
TreeNode 結構

class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;} }

Recursive

class Solution {private int sum = 0;public int sumOfLeftLeaves(TreeNode root) {if (root == null) {return 0;}if (root.left != null && root.left.left == null && root.left.right == null) {sum += root.left.val;}sumOfLeftLeaves(root.left);sumOfLeftLeaves(root.right);return sum;} } class Solution {public int sumOfLeftLeaves(TreeNode root) {int count = 0;if (root == null) {return 0;}if (root.left != null) {if (root.left.left == null && root.left.right == null) {count += root.left.val;} else {count += sumOfLeftLeaves(root.left);}}count += sumOfLeftLeaves(root.right);return count;} }

Iterative

import java.util.Stack; class Solution {public int sumOfLeftLeaves(TreeNode root) {int count = 0;Stack<TreeNode> stack = new Stack<>();if (root == null) {return 0;}stack.push(root);while (!stack.isEmpty()) {TreeNode node = stack.pop();if (node.left != null) {if (node.left.left == null && node.left.right == null) {count += node.left.val;} else {stack.push(node.left);}}if (node.right != null) {if (node.right.left != null || node.right.right != null) {stack.push(node.right);}}}return count;} }

主測試類

public class Test {public static void main(String[] args) {Solution tree = new Solution();/* create a tree */TreeNode root = new TreeNode(3);root.left = new TreeNode(9);root.right = new TreeNode(20);root.right.left = new TreeNode(15);root.right.right = new TreeNode(7);System.out.println(tree.sumOfLeftLeaves(root));} }

運行結果

24

相似題目

• 求二叉樹中葉子節點的個數

參考資料

• https://leetcode.com/problems/sum-of-left-leaves/
• https://leetcode-cn.com/problems/sum-of-left-leaves/

轉載于:https://www.cnblogs.com/hglibin/p/10849527.html

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