Leetcode——二叉树中和为某一值的路径（路径总和 II）

Posted 2021-08-29 Yawn,

1. 题目

2. 题解

（1）DFS

• 深度优先遍历每一个路径，符合条件的存下来

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    //深度优先遍历(回溯)
    List<List<Integer>> res = new ArrayList<>();
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        if(root == null)
            return res;
        dfs(new ArrayList<Integer>(), 0, root, targetSum);
        return res;
    }

    public void dfs(ArrayList<Integer> curPath, int curSum, TreeNode node, int targetSum ){
        //能进来肯定保证非空
        curPath.add(node.val);
        curSum+=node.val;

        //判断符合条件的路径
        if(node.left == null && node.right == null && curSum == targetSum){
            res.add(new ArrayList<>(curPath));
        }

        if(node.left != null)           
            dfs(curPath,curSum,node.left,targetSum);

        if(node.right != null)        
            dfs(curPath,curSum,node.right,targetSum);
        
        curPath.remove(curPath.size()-1);   //回溯,移除最后一位不符合条件的值
        //curSum-=node.val;            ：不需要这句，每个节点都有针对他自己的curSum，不需要回溯
    }
}

（2）BFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    List<List<Integer>> ret = new LinkedList<List<Integer>>();
    Map<TreeNode, TreeNode> map = new HashMap<TreeNode, TreeNode>();    //保存，左子树跟右子树的根节点

    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        if (root == null) {
            return ret;
        }

        Queue<TreeNode> queueNode = new LinkedList<TreeNode>();
        Queue<Integer> queueSum = new LinkedList<Integer>();    //记录当前节点的和
        queueNode.offer(root);
        queueSum.offer(0);

        while (!queueNode.isEmpty()) {
            TreeNode node = queueNode.poll();
            int rec = queueSum.poll() + node.val;

            if (node.left == null && node.right == null) {
                if (rec == targetSum) {
                    getPath(node);
                }
            } else {
                if (node.left != null) {
                    map.put(node.left, node);
                    queueNode.offer(node.left);
                    queueSum.offer(rec);
                }
                if (node.right != null) {
                    map.put(node.right, node);
                    queueNode.offer(node.right);
                    queueSum.offer(rec);
                }
            }
        }

        return ret;
    }

    //根据Map存的父节点子，回溯找到路径，注意：此时找到路径是反的
    public void getPath(TreeNode node) {
        List<Integer> temp = new LinkedList<Integer>();
        while (node != null) {
            temp.add(node.val);			//可以使用temp.add(0, node.val);  
            node = map.get(node);
        }
        Collections.reverse(temp);		//注释掉这句即可
        ret.add(new LinkedList<Integer>(temp));
    }
}