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Leetcode 104 Maximum Depth Of Binary Tree

发表于 更新于 分类于 Valine：
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Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Note: A leaf is a node with no children.

Example:

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

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2
3
4
5
  3
 / \
9  20
  /  \
 15   7

return its depth = 3.

1 递归

直接递归非常简单

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/**
 * 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 maxDepth(TreeNode root) {
            return root == null ? 0 : Math.max(maxDepth(root.left) + 1, maxDepth(root.right) + 1);
        }
}

2 BFS

每一次将一层的节点(非空)加入队列, 看看一共加入了多少次.

比递归稍慢.

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/**
 * 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 maxDepth(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<>();
        if(root == null)
            return 0;
        q.offer(root);
        int depth = 0;
        while(!q.isEmpty())
        {
            int size = q.size();
            while(size != 0)
            {
                TreeNode tmp = q.poll();
                if(tmp.left != null)
                    q.offer(tmp.left);
                if(tmp.right != null)
                    q.offer(tmp.right);
                --size;
            }
            ++depth;
        }
        return depth;
    }
}