🌗 111. 二叉树的最小深度

难度: 🌗

问题描述

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解法 1 - 递归

/**
 * 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 {
    public int minDepth(TreeNode root) {
        // 思路：
        // 叶子节点，即左右子节点均空的节点
        return mySol(root);
    }

    private int mySol(TreeNode root) {
        // 递归终止条件
        if(root == null) {
            return 0;
        }
        if(root.left == null && root.right == null) {
            return 1;
        }
        if(root.left == null) {
            return mySol(root.right) + 1;
        }
        if(root.right == null) {
            return mySol(root.left) + 1;
        }
        return Math.min(mySol(root.left), mySol(root.right)) + 1;
    }
}

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输出 1

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解法 2 - 层序遍历

/**
 * 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 {
    public int minDepth(TreeNode root) {
        // 思路：
        // 层序 - 辅助 队列
        int res = 0;
        if(root == null) {
            return res;
        }
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            res ++;
            int len = queue.size();
            for(int i = 0; i < len; i ++) {
                TreeNode cur = queue.poll();
                if(cur.left == null && cur.right == null) {
                    return res;
                }
                if(cur.left != null) {
                    queue.offer(cur.left);
                }
                if(cur.right != null) {
                    queue.offer(cur.right);
                }
            }
        }
        return res;
    }
}

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输出 2