1.题目描述
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.
2.解决方案1class Solution {public:int maxDepth(TreeNode *root) {if(root == NULL){return 0;}deque<TreeNode*> deqNodes;deqNodes.push_back(root);int depth = 0;int alreadyDoneCount = 1;while(deqNodes.empty() == false){TreeNode* lastNode = deqNodes.back();deqNodes.pop_back();alreadyDoneCount–;if(lastNode->left){deqNodes.push_front(lastNode->left);}if(lastNode->right){deqNodes.push_front(lastNode->right);}if(alreadyDoneCount == 0){++depth;alreadyDoneCount = deqNodes.size();}}return depth;}};思路：采用宽度优先搜索，只有当添加的全处理了，，树的深度才增加1，这个是个小技巧。
3.解决方案2
class Solution {public:int maxDepth(TreeNode *root) {if(root == NULL ){return 0;}else if(root != NULL && (root->left == NULL && root->right == NULL)){return 1;}int leftTreeDepth = maxDepth(root->left);int rightTreeDepth = maxDepth(root->right);int maxDepth = leftTreeDepth > rightTreeDepth ? leftTreeDepth : rightTreeDepth;return maxDepth + 1;}};
思路：树的遍历当然可以用递归，但会比较慢。
一错再错，把握正确的方向，
