【LeetCode】 33 路径总和

题目

给定一个二叉树和一个目标和,判断该树中是否存在根节点到叶子节点的路径,这条路径上所有节点值相加等于目标和。

说明: 叶子节点是指没有子节点的节点。

示例:
给定如下二叉树,以及目标和 sum = 22,

          5
         / \
        4   8
       /   / \
      11  13  4
     /  \      \
    7    2      1

返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。

思路

https://leetcode-cn.com/problems/path-sum/solution/lu-jing-zong-he-by-leetcode/

代码

递归

class Solution {
  public boolean hasPathSum(TreeNode root, int sum) {
    if (root == null)
      return false;

    sum -= root.val;
    if ((root.left == null) && (root.right == null))
      return (sum == 0);
    return hasPathSum(root.left, sum) || hasPathSum(root.right, sum);
  }
}

迭代

class Solution {
  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);

    TreeNode node;
    int curr_sum;
    while ( !node_stack.isEmpty() ) {
      node = node_stack.pollLast();
      curr_sum = sum_stack.pollLast();
      if ((node.right == null) && (node.left == null) && (curr_sum == 0))
        return true;

      if (node.right != null) {
        node_stack.add(node.right);
        sum_stack.add(curr_sum - node.right.val);
      }
      if (node.left != null) {
        node_stack.add(node.left);
        sum_stack.add(curr_sum - node.left.val);
      }
    }
    return false;
  }
}

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