力扣每日一题，113. 路径总和 II

我的成绩

看这成绩，显然代码还需优化。不过我觉都我这种半吊子能过就很不错了，优化什么的，暂时去你的吧。

题目描述

给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

示例:
给定如下二叉树，以及目标和 sum = 22，

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

返回:

[
   [5,4,11,2],
   [5,8,4,5]
]

思路

总之二叉树的题，就遍历了再说，十有八九是需要遍历的。

遍历模板

递归实现

public class Solution {
    public static void main(String[] args) {
        /*
         *       5
         *      / \
         *     4   3
         *    / \
         *   1   2
         */
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(4);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(1);
        root.left.right = new TreeNode(2);
        new Solution().preOrder(root);
        // 递归实现
    }
    void preOrder(TreeNode root) {
    	if (root != null) {
    		System.out.print(root.val + " ");
    		preOrder(root.left);
    		preOrder(root.right);
    	}
    }
}
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) {
        val = x;
    }
}

提交代码

package 路径总和II;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;

import org.junit.Test;

class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;

	TreeNode(int x) {
		val = x;
	}
}

public class Solution {
	@Test
	public void testPathSum() throws Exception {
		TreeNode root = new TreeNode(5);
		root.left = new TreeNode(4);
		root.right = new TreeNode(8);
		root.left.left = new TreeNode(11);
		root.left.left.left = new TreeNode(7);
		root.left.left.right = new TreeNode(2);

		root.right.left = new TreeNode(13);
		root.right.right = new TreeNode(4);
		root.right.right.left = new TreeNode(5);
		root.right.right.right = new TreeNode(1);

		pathSum(root, 22);

	}

	public List<List<Integer>> pathSum(TreeNode root, int sum) {
		List<Integer> path = new LinkedList<Integer>();
		List<List<Integer>> res = new LinkedList<List<Integer>>();
		preOrder(root, 0, sum, path, res);
		return res;
	}

	private void preOrder(TreeNode root,int num,int sum,List<Integer> path,List<List<Integer>> res) {
		if(root!=null){
			num+=root.val;
			path.add(root.val);
			if(root.left==null&&root.right==null&&num==sum){
				System.out.println(path);
				res.add(new ArrayList<Integer>(path));
			}
//			System.out.println(root.val);
			preOrder(root.left, num,sum, path, res);
			preOrder(root.right, num,sum, path, res);
			path.remove(path.size()-1);
			num-=root.val;
		}

	}
}