2015年9月9日星期三

Lintcode: Maximum Depth of Binary Tree


My leetcode same problem solution

1. Recursion(DFS):
n -- number of nodes
L -- height of tree
Time: O(n)
Space: O(h)
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/**
 * 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) {
        // write your code here
        if(!root) return 0;
        int left = 1;
        int right = 1;
        if(root->left) left = maxDepth(root->left) + left;
        if(root->right) right = maxDepth(root->right) + right;
        return (left>right)?left:right;
    }
   
};

2. Iterative solution(BFS):
The deepest node is located at the end of the queue
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/**
 * 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) {
        queue<pair<TreeNode*, int> > q;
        if(!root) return 0;
        q.push(make_pair(root, 1));
        while(!q.empty()){
            pair<TreeNode*, int> cur = q.front();
            q.pop();
            if(cur.first->left) q.push(make_pair(cur.first->left, cur.second+1) );
            if(cur.first->right) q.push(make_pair(cur.first->right, cur.second+1) );
            if(q.empty()) return cur.second;
        }
    }
   
};
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