题目113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/
4 8
/ /
11 13 4
/ \ /
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]

思路: 参考112. Path Sum

1,递归

public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> results = new ArrayList<List<Integer>>();
        List<Integer> result = new ArrayList<Integer>();
        dfs(root,0,sum,result,results);
        return results;
    }
    
    private void dfs(TreeNode root, int curSum, int sum, List<Integer> result, List<List<Integer>> results){
         if(root == null){
            return;
         }
         
         if(root.left == null && root.right == null){
            if (curSum + root.val == sum){
                result.add(root.val);
                results.add(result);
            }
            return;
         }
         
         result.add(root.val);
         if(root.left != null && root.right != null){
             List<Integer> tempResult = new ArrayList<Integer>();
             tempResult.addAll(result);
             dfs(root.left,curSum + root.val,sum,tempResult,results); 
             dfs(root.right,curSum + root.val,sum,result,results); 
         }else if(root.left != null && root.right == null){
             dfs(root.left,curSum + root.val,sum,result,results);
         }else if(root.left == null && root.right != null){
             dfs(root.right,curSum + root.val,sum,result,results);
         }
     }

2,利用后序遍历

public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if(root == null){
            return result;
        }
        
          //根节点到栈顶节点的路径
        Stack<TreeNode> pathsStack = new Stack<TreeNode>();
        TreeNode node = root;
        TreeNode tempNode = null;
        boolean hasVisited = true;
        do{
            while(node != null){
              pathsStack.push(node);
              node = node.left;
            }
            
            tempNode = null;
            hasVisited = true;
            while(!pathsStack.empty() && hasVisited){
                node = pathsStack.peek();
                if(tempNode == node.right){
                    tempNode = pathsStack.pop();
                    //判断根节点到叶子节点的路径和
                    if(tempNode.left == null && tempNode.right == null){
                        int pathSum = tempNode.val;
                        List<Integer> path = new LinkedList<Integer>();
                        for(TreeNode treeNode : pathsStack){
                            pathSum += treeNode.val;
                            path.add(treeNode.val);
                        }
                    
                        if(sum == pathSum){
                            result.add(path);
                            path.add(tempNode.val);
                        }
                    }
                }else{
                    node = node.right;
                    hasVisited = false;
                }
            }
            
        }while(!pathsStack.empty());
        return result;
    }