113_路径总和II

113_路径总和II 

 

package 二叉树.BT;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Deque;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Queue;
/**
 * https://leetcode-cn.com/problems/path-sum-ii/
 * @author Huangyujun
 *
 */
public class _113_路径总和II {

    class Solution {
        List<List<Integer>> ret = new LinkedList<List<Integer>>();
        Deque<Integer> path = new LinkedList<Integer>();

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

        public void dfs(TreeNode root, int targetSum) {
            if (root == null) {
                return;
            }
            path.offerLast(root.val);
            targetSum -= root.val;
            if (root.left == null && root.right == null && targetSum == 0) {
                ret.add(new LinkedList<Integer>(path));
            }
            dfs(root.left, targetSum);// 先到这里的话是叶子结点【可能是叶子（左边已经遍历完）】
            dfs(root.right, targetSum);// 然后叶子结点的右边
            path.pollLast(); // 这里poll掉我有疑惑？？？-------右边遍历完，必然是叶子
        }
    }

    /**
     * [可能是使用了map 集合，效率很低哈哈哈哈]
     * 思路分析：
     * map 结合【键：当前结点， 值：父结点】
     * 队列queueSum 是计算路径的值【将计算过结果放到栈顶】，stack 遍历结点的栈 
     * @author Huangyujun
     *
     */
    class Solution2 {
        List<List<Integer>> ret = new LinkedList<List<Integer>>();
        Map<TreeNode, TreeNode> map = new HashMap<TreeNode, TreeNode>();

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

            Queue<TreeNode> stack = new LinkedList<TreeNode>();
            Queue<Integer> queueSum = new LinkedList<Integer>();
            stack.offer(root);
            queueSum.offer(0);

            while (!stack.isEmpty()) {
                TreeNode node = stack.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);
                        stack.offer(node.left);
                        queueSum.offer(rec);
                    }
                    if (node.right != null) {
                        map.put(node.right, node);
                        stack.offer(node.right);
                        queueSum.offer(rec);
                    }
                }
            }

            return ret;
        }

        public void getPath(TreeNode node) {
            List<Integer> temp = new LinkedList<Integer>();
            while (node != null) {
                temp.add(node.val);
                node = map.get(node);
            }
            Collections.reverse(temp);
            ret.add(new LinkedList<Integer>(temp));
        }
    }

    /**
     * 思路不对，我只想明白了左到尽头不一定是叶子的bug
     *后边进行判断了，为啥还是bug
     * 
     * @param root
     * @param targetSum
     * @return
     */
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<Integer> item = new LinkedList<>();
        List<List<Integer>> res = new LinkedList<>();
        Deque<TreeNode> stack = new LinkedList<>();
        TreeNode node = root;
        int num = 0;
        while (node != null || !stack.isEmpty()) {
            while (node != null) {
                item.add(node.val);
                num += node.val;
                stack.push(node);
                node = node.left;
            }
            // 左到尽头，不是叶子，可能度为1【需要进行判断】
            // 判断一下
            if (stack.peek().right == null) {
                if (num == targetSum) {
                    res.add(item);
                }
                // 到达叶子了
                int temp = ((Deque<Integer>) item).pollLast();
                num -= temp;
            }
            // 吐掉当前结点
            node = stack.pop();
            node = node.right;
        }
        return res;
    }
}