【Leetcode】Maximum Depth of Binary Tree

最新推荐文章于 2025-08-20 17:55:57 发布

转载最新推荐文章于 2025-08-20 17:55:57 发布 · 191 阅读

文章标签：

#java #leetcode #binary tree

本文介绍了两种求解二叉树最大深度的方法：递归法和迭代法。递归法通过深度优先搜索来确定从根节点到最远叶节点的最长路径上的节点数。迭代法则利用队列实现广度优先搜索，逐层遍历树的所有节点，直至遍历完整棵树。

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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.

Java:

Recursion

http://blog.youkuaiyun.com/u010119170/article/details/24739435

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int maxDepth(TreeNode root) {
        if (root==null)
        {
            return 0;
        }
        int leftDepth=maxDepth(root.left);
        int rightDepth=maxDepth(root.right);
        return leftDepth > rightDepth ? leftDepth + 1 : rightDepth + 1;
        //if leftDepth is larger, left Depth +1; 
    }
}

Iteration

水印人生：http://gongxuns.blogspot.com/2012/12/leetcodemaximum-depth-of-binary-tree.html

public class Solution {
    public int maxDepth(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<TreeNode> prev = new ArrayList<TreeNode>();
        
        if(root!=null) prev.add(root);
        int depth=0;
        while(!prev.isEmpty()){
            ArrayList<TreeNode> temp = new ArrayList<TreeNode> ();
            for(TreeNode node:prev){
                if(node.left!=null) temp.add(node.left);
                if(node.right!=null) temp.add(node.right);
            }    
            prev = new ArrayList<TreeNode>(temp);
            depth++;
        }
        return depth;
    }
}

http://blog.youkuaiyun.com/linhuanmars/article/details/19659525?reload

public int maxDepth(TreeNode root) {
    if(root == null)
        return 0;
    int level = 0;
    LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
    queue.add(root);
    int curNum = 1; //num of nodes left in current level
    int nextNum = 0; //num of nodes in next level
    while(!queue.isEmpty())
    {
        TreeNode n = queue.poll();
        curNum--;
        if(n.left!=null)
        {
            queue.add(n.left);
            nextNum++;
        }
        if(n.right!=null)
        {
            queue.add(n.right);
            nextNum++;
        }
        if(curNum == 0)
        {
            curNum = nextNum;
            nextNum = 0;
            level++;
        }
    }
    return level;
}