leetcode | 112. Path Sum(递归&&迭代)

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本文详细解析了LeetCode上路径总和问题的两种解决方案：递归法和迭代法。通过实例展示如何在给定的二叉树中寻找根到叶子节点路径，其节点值之和等于指定值。第一种方法采用递归策略，简洁直观；第二种方法使用迭代，通过栈实现深度优先搜索。两种方法的时间复杂度均为O(n)，空间复杂度平均为O(log(n))，最坏情况下为O(n)。

Problem

https://leetcode.com/problems/path-sum/
Example:

Given the below binary tree and sum = 22,
5
/
4 8
/ /
11 13 4
/ \
7 2 1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Solution1

参见 https://leetcode.com/problems/path-sum/solution/

递归
time: O(n)
space: average O(log(n)) worst O(n)

    public boolean hasPathSum(TreeNode root, int sum) {
        if(root==null) return false;
        if(root.left==null&&root.right==null) return root.val == sum;
        return hasPathSum(root.left, sum - root.val) ||  hasPathSum(root.right, sum - root.val);
    }

Solution2

迭代，dfs
time: O(n)
space: average O(log(n)) worst O(n)

public boolean hasPathSum(TreeNode root, int sum) {
        if(root==null) return false;
        LinkedList<TreeNode> node_stack = new LinkedList();
        LinkedList<Integer> sum_stack = new LinkedList();
        node_stack.add(root);
        sum_stack.add(sum - root.val);
        while(!node_stack.isEmpty()){
            TreeNode node = node_stack.pollLast();
            int tmp = sum_stack.pollLast();
            if(node.left==null&&node.right==null&&tmp==0) return true;
            if(node.left!=null){
                node_stack.add(node.left);
                sum_stack.add(tmp - node.left.val);
            }
            if(node.right!=null){
                node_stack.add(node.right);
                sum_stack.add(tmp - node.right.val);
            }
        }
        return false;
    }