LeetCode_104_Maximum Depth of Binary Tree

104. Maximum Depth of Binary Tree

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.

 

题目分析:

    给定一个二叉树,求出他的最大深度。

    采用递归的方法,不断求解l,r的深度,当l or r = NULL时 返回 0, 每次调用递归函数时深度+1。并不断取l,r的最大值。

 

Solution

/**
 * Definition for a binary tree node.
 * 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 l = maxDepth(root->left);
        int r = maxDepth(root->right);
        
        return max(l,r)+1;
    }
};

 

 

对于递归掌握不是特别熟练,验证思路

LeetCode_104_Maximum Depth of Binary Tree 

下面以(‘x’)表示值为x的结点(仅针对上图以方便表达)

  maxDepth(‘5’) = max( maxDepth(‘4’), maxDepth(‘7’) )+1

  maxDepth(‘4’) = max( maxDepth(‘3’), 0 )+1

  maxDepth(‘7’) = max( maxDepth(‘2’), 0 )+1

  maxDepth(‘3’) = max( maxDepth(‘-1’), 0 )+1

  maxDepth(‘2’) = max( 0 ,0 )+1

  maxDepth(‘-1’) = max( 0, 0 )+1

 

所以

  maxDepth(‘-1’) = 1;

  maxDepth(‘2’) = 1;

  maxDepth(‘3’) = 2;

  maxDepth(‘7’) = 2;

  maxDepth(‘4’) = 3;

  maxDepth(‘5’) = 4;

分解为各个子树,深度求解亦正确。