二叉树的最大深度和最小深度(编程实现)

1. 二叉树的最大深度:(LeetCode104)  

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.

最大深度也是到叶子节点的长度,但是因为是求最大深度,单个孩子为空的非叶子节点不会干扰到结果,因此用最简洁的处理方式就可以搞定。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode *root) {
        if(!root) return 0;
        int leftDepth = maxDepth(root -> left) + 1;      // 递归调用左子树
        int rightDepth = maxDepth(root -> right) + 1;    // 递归调用右子树
        return max(leftDepth, rightDepth);               // 返回左右子树中较大的一个,  注意最后不用加 1。
    }
};

2. 二叉树的最小深度(LeetCode111)

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.

因为深度是必须到叶子节点的距离,因此使用深度遍历时,不能单纯的比较左右子树的递归结果返回较小值,因为对于有单个孩子为空的节点,为空的孩子会返回0,但这个节点并非叶子节点,故返回的结果是错误的。因此,当发现当前处理的节点有单个孩子是空时,返回一个极大值INT_MAX,防止其干扰结果。

1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
 class Solution {
  public:
      int minDepth(TreeNode *root) {
          if(!root) return 0;
          if(!root -> left && !root -> right) return 1;   //Leaf means should return depth.
          int leftDepth = 1 + minDepth(root -> left);
          leftDepth = (leftDepth == 1 ? INT_MAX : leftDepth);
          int rightDepth = 1 + minDepth(root -> right);
          rightDepth = (rightDepth == 1 ? INT_MAX : rightDepth);  //If only one child returns 1, means this is not leaf, it does not return depth.
         return min(leftDepth, rightDepth);
     }
 };
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode *root) {
        if (root == NULL) return 0;
        if (root->left == NULL && root->right == NULL) return 1;
        
        if (root->left == NULL) return minDepth(root->right) + 1;
        else if (root->right == NULL) return minDepth(root->left) + 1;
        else return 1 + min(minDepth(root->left), minDepth(root->right));
    }
    
};
原文地址:https://www.cnblogs.com/simplepaul/p/7674586.html