路径总和Ⅱ

113. 路径总和ii

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：[]

示例 3：

输入：root = [1,2], targetSum = 0
输出：[]

提示：

• 树中节点总数在范围 [0, 5000] 内
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

回溯

/**
 * 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>> ans = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        if(root == null) return ans;
        traversal(root, targetSum, root.val);
        return ans;
    }


    public void traversal(TreeNode root, int targetSum, int sum){
        path.add(root.val);
        if(sum == targetSum && root.left == null && root.right == null){
            ans.add(new ArrayList<>(path));
        }
        if(root.left != null){
            traversal(root.left, targetSum, sum + root.left.val);
            path.remove(path.size()-1);
        }
        if(root.right != null){
            traversal(root.right, targetSum, sum + root.right.val);
            path.remove(path.size()-1);
        }

    }
}

迭代（后序遍历）

class Solution {

    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<List<Integer>> ans = new ArrayList<>(); //保存答案
        List<Integer> path = new ArrayList<>();//保存当前路径

         Stack<TreeNode> st = new Stack<>();
         TreeNode pre = null;
         //后序遍历
         while(root != null || !st.isEmpty()){
             while(root != null){
                 st.push(root);
                 root = root.left;
             }
             root = st.peek();
             if(root.right == null || pre == root.right){//右节点未被访问或为空，访问该节点
                if(root.left == null && root.right ==null){//遍历到叶子节点
                    int pathSum = 0; //记录当前路径的总和
                    path.clear();
                    for(TreeNode node : st){
                        pathSum += node.val;
                        path.add(node.val);
                    }

                    if(pathSum == targetSum){ //路径和等于目标和，加入该路径
                        ans.add(new ArrayList<>(path));
                    }
                }
                root = st.pop(); //访问节点出栈
                pre = root;
                root = null;
             }else{
                 root = root.right;
             }
        }
        return ans;
    }

}