Question

Source

• leetcode: Maximum Depth of Binary Tree | LeetCode OJ
• lintcode: (97) Maximum Depth of Binary Tree

Problem

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.

Example

Given a binary tree as follow:

  1
 / \
2   3
   / \
  4   5

The maximum depth is 3 .

题解 - 递归

树遍历的题最方便的写法自然是递归，不过递归调用的层数过多可能会导致栈空间溢出，因此需要适当考虑递归调用的层数。我们首先来看看使用递归如何解这道题，要求二叉树的最大深度，直观上来讲使用深度优先搜索判断左右子树的深度孰大孰小即可，从根节点往下一层树的深度即自增 1，遇到 NULL 时即返回 0。

由于对每个节点都会使用一次 maxDepth ，故时间复杂度为 O(n)O(n)O(n), 树的深度最大为 nnn, 最小为 log2n\log_2 nlog2n, 故空间复杂度介于 O(logn)O(\log n)O(logn) 和 O(n)O(n)O(n) 之间。

C++

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *   int val;
 *   TreeNode *left, *right;
 *   TreeNode(int val) {
 *     this->val = val;
 *     this->left = this->right = NULL;
 *   }
 * }
 */
class Solution {
public:
  /**
   * @param root: The root of binary tree.
   * @return: An integer
   */
  int maxDepth(TreeNode *root) {
    if (NULL == root) {
      return 0;
    }

    int left_depth = maxDepth(root->left);
    int right_depth = maxDepth(root->right);

    return max(left_depth, right_depth) + 1;
  }
};

Java

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left, right;
 *   public TreeNode(int val) {
 *     this.val = val;
 *     this.left = this.right = null;
 *   }
 * }
 */
public class Solution {
  /**
   * @param root: The root of binary tree.
   * @return: An integer.
   */
  public int maxDepth(TreeNode root) {
    // write your code here
    if (root == null) {
      return 0;
    }
    return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
  }
}

题解 - 迭代（显式栈)

使用递归可能会导致栈空间溢出，这里使用显式栈空间（使用堆内存) 来代替之前的隐式栈空间。从上节递归版的代码（先处理左子树，后处理右子树，最后返回其中的较大值) 来看，是可以使用类似后序遍历的迭代思想去实现的。

首先使用后序遍历的模板，在每次迭代循环结束处比较栈当前的大小和当前最大值 max_depth 进行比较。

C++

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *   int val;
 *   TreeNode *left, *right;
 *   TreeNode(int val) {
 *     this->val = val;
 *     this->left = this->right = NULL;
 *   }
 * }
 */
class Solution {
public:
  /**
   * @param root: The root of binary tree.
   * @return: An integer
   */
  int maxDepth(TreeNode *root) {
    if (NULL == root) {
      return 0;
    }

    TreeNode *curr = NULL, *prev = NULL;
    stack<TreeNode *> s;
    s.push(root);

    int max_depth = 0;

    while(!s.empty()) {
      curr = s.top();
      if (!prev || prev->left == curr || prev->right == curr) {
        if (curr->left) {
          s.push(curr->left);
        } else if (curr->right){
          s.push(curr->right);
        }
      } else if (curr->left == prev) {
        if (curr->right) {
          s.push(curr->right);
        }
      } else {
        s.pop();
      }

      prev = curr;

      if (s.size() > max_depth) {
        max_depth = s.size();
      }
    }

    return max_depth;
  }
};

题解 3 - 迭代（队列)

在使用了递归/后序遍历求解树最大深度之后，我们还可以直接从问题出发进行分析，树的最大深度即为广度优先搜索中的层数，故可以直接使用广度优先搜索求出最大深度。

C++

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *   int val;
 *   TreeNode *left, *right;
 *   TreeNode(int val) {
 *     this->val = val;
 *     this->left = this->right = NULL;
 *   }
 * }
 */
class Solution {
public:
  /**
   * @param root: The root of binary tree.
   * @return: An integer
   */
  int maxDepth(TreeNode *root) {
    if (NULL == root) {
      return 0;
    }

    queue<TreeNode *> q;
    q.push(root);

    int max_depth = 0;
    while(!q.empty()) {
      int size = q.size();
      for (int i = 0; i != size; ++i) {
        TreeNode *node = q.front();
        q.pop();

        if (node->left) {
          q.push(node->left);
        }
        if (node->right) {
          q.push(node->right);
        }
      }

      ++max_depth;
    }

    return max_depth;
  }
};

Java

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left, right;
 *   public TreeNode(int val) {
 *     this.val = val;
 *     this.left = this.right = null;
 *   }
 * }
 */
public class Solution {
  /**
   * @param root: The root of binary tree.
   * @return: An integer.
   */
  public int maxDepth(TreeNode root) {
    if (root == null) {
      return 0;
    }

    int depth = 0;
    Queue<TreeNode> q = new LinkedList<TreeNode>();
    q.offer(root);
    while (!q.isEmpty()) {
      depth++;
      int qLen = q.size();
      for (int i = 0; i < qLen; i++) {
        TreeNode node = q.poll();
        if (node.left != null) q.offer(node.left);
        if (node.right != null) q.offer(node.right);
      }
    }

    return depth;
  }
}

源码分析

广度优先中队列的使用中， qLen 需要在 for 循环遍历之前获得，因为它是一个变量。

复杂度分析

最坏情况下空间复杂度为 O(n)O(n)O(n), 遍历每一个节点，时间复杂度为 O(n)O(n)O(n),