leetcode 广搜小题两道

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mb6304a73bed12a 2022-08-23 19:33:28 博主文章分类：LeetCode ©著作权

文章标签 广搜 ide 最小值 文章分类 虚拟化云计算

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leetcode 广搜小题两道

周末无聊，故刷两道小题，用的都是广搜的思想

111. Minimum Depth of Binary Tree

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

3 / \ 9 20 / \ 15 7

return its minimum depth = 2.

求树的最小深度。思路是用广搜，按层遍历树，当遇到叶子节点时，返回当前的层数。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        if(root == null) {
            return 0;
        }
     
        LinkedList<TreeNode> list = new LinkedList<>();
        list.add(root);
        int depth = 0;
     
        while(list.size() > 0) {
            int length = list.size();
            depth++;
            for(int i = 0; i < length; i++) {
                TreeNode node = list.removeFirst();
                if(node.left == null && node.right == null) {
                    return depth;
                }
                if(node.left != null) {
                    list.add(node.left);
                }
                if(node.right != null) {
                    list.add(node.right);
                }    
            }
        }
        return depth;
    }
}

112. Second Minimum Node In a Binary Tree

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or zero sub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes.

second minimum

If no such second minimum value exists, output -1 instead.

Example 1:

Input:

Output: 5
Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input: 2 / \ 2 2 Output: -1 Explanation:

这道题的二叉树有一个的特点：父节点的值，是左、右孩子的最小值。由此推断出：根节点是整棵树中的最小值，即树中第一小的值。要找到树中第二小的值，只需要找到那些比根节点值大的节点，并从中找出拥有最小值的节点。

依然用广搜的思想，按层遍历树，如果遇到一个节点的值大于根节点，说明这个节点的左、右子树中的所有节点的值，都大于根节点值，我们就不用遍历该节点的孩子们了。

比如，Example 1中，节点5大于树根2，则节点5的左、右孩子也都大于树根2。既然我们现在要找到树中第二小的值，那肯定不会是5的左右孩子了。第二小的值，应该是一个范围在**(2,5]**之间的值。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int findSecondMinimumValue(TreeNode root) {
        LinkedList<TreeNode> list = new LinkedList<>();
        list.add(root);
        int min = Integer.MAX_VALUE;
        
        while(list.size() > 0) {
            TreeNode node = list.removeFirst();            
            if(node.val > root.val) {
                min = Math.min(min, node.val);
                continue;
            }
            if(node.left != null) {
                list.add(node.left);
            }
            if(node.right != null) {
                list.add(node.right);
            }            
        }
        return min == Integer.MAX_VALUE ? -1 : min;
    }
}

199. Binary Tree Right Side View

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

Example:

Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:

1 <—
/ \
2 3 <—
\ \
5 4 <—

题目大意：
右视图下的二叉树，从右边放眼望去，输出看到的节点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new LinkedList<>();
        if(root == null) {
            return result;
        }
        LinkedList<TreeNode> list = new LinkedList<>();
        list.add(root);
        while(!list.isEmpty()) {
            int range = list.size();
            for(int i = 0; i < range; i++) {
                TreeNode node = list.removeFirst();
                if(node.left != null) {
                    list.add(node.left);
                }
                if(node.right != null) {
                    list.add(node.right);
                }
                // 把最右边的节点加入到result中
                if(i == range - 1) {
                    result.add(node.val);
                }
            }
        }
        return result;
    }
}