[剑指Offer] 38.二叉树的深度

题目描述

输入一棵二叉树，求该树的深度。从根结点到叶结点依次经过的结点（含根、叶结点）形成树的一条路径，最长路径的长度为树的深度。

【思路1】递归

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
public:
    int TreeDepth(TreeNode* pRoot)
    {
        if(pRoot == NULL)
            return 0;
        int h1 = TreeDepth(pRoot->left);
        int h2 = TreeDepth(pRoot->right);
        return max(h1,h2) + 1;
    }
};

 【思路2】DFS，用一个栈来存储结点，一个栈来存储当前深度

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution
{
public:
    int TreeDepth(TreeNode* pRoot)
    {
        if(pRoot == NULL) return 0;
        stack<TreeNode*> nodeStack;
        stack<int> depthStack;
        
        nodeStack.push(pRoot);
        depthStack.push(1);
        
        TreeNode* node = pRoot;
        int MaxDepth = 0;
        int curDepth = 0;
        
        while(!nodeStack.empty())
        {
            node = nodeStack.top();
            nodeStack.pop();
            curDepth = depthStack.top();
            depthStack.pop();
            if(MaxDepth < curDepth)
                MaxDepth = curDepth;
            
            if(node->left != NULL)
            {
                nodeStack.push(node->left);
                depthStack.push(curDepth + 1);
            }
            if(node->right != NULL)
            {
                nodeStack.push(node->right);
                depthStack.push(curDepth + 1);
            }
        }
        return MaxDepth;
    }
};