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# Maximum Depth of Binary Tree ### Source - lintcode: [(97) Maximum Depth of Binary Tree](http://www.lintcode.com/en/problem/maximum-depth-of-binary-tree/) ### Problem Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the rootnode down to the farthest leaf node. #### Example Given a binary tree as follow: ~~~ 1 / \ 2 3 / \ 4 5 ~~~ The maximum depth is `3`. ### 题解 - 递归 树遍历的题最方便的写法自然是递归,不过递归调用的层数过多可能会导致栈空间溢出,因此需要适当考虑递归调用的层数。我们首先来看看使用递归如何解这道题,要求二叉树的最大深度,直观上来讲使用深度优先搜索判断左右子树的深度孰大孰小即可,从根节点往下一层树的深度即自增1,遇到`NULL`时即返回0。 由于对每个节点都会使用一次`maxDepth`,故时间复杂度为 O(n)O(n)O(n), 树的深度最大为 nnn, 最小为 log2n\log_2 nlog2n, 故空间复杂度介于 O(logn)O(\log n)O(logn) 和 O(n)O(n)O(n) 之间。 ### C++ ~~~ /** * 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) { if (NULL == root) { return 0; } int left_depth = maxDepth(root->left); int right_depth = maxDepth(root->right); return max(left_depth, right_depth) + 1; } }; ~~~ ### Java ~~~ /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: The root of binary tree. * @return: An integer. */ public int maxDepth(TreeNode root) { // write your code here if (root == null) { return 0; } return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1; } } ~~~ ### 题解 - 迭代(显式栈) 使用递归可能会导致栈空间溢出,这里使用显式栈空间(使用堆内存)来代替之前的隐式栈空间。从上节递归版的代码(先处理左子树,后处理右子树,最后返回其中的较大值)来看,是可以使用类似后序遍历的迭代思想去实现的。 首先使用后序遍历的模板,在每次迭代循环结束处比较栈当前的大小和当前最大值`max_depth`进行比较。 ### C++ ~~~ /** * 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) { if (NULL == root) { return 0; } TreeNode *curr = NULL, *prev = NULL; stack<TreeNode *> s; s.push(root); int max_depth = 0; while(!s.empty()) { curr = s.top(); if (!prev || prev->left == curr || prev->right == curr) { if (curr->left) { s.push(curr->left); } else if (curr->right){ s.push(curr->right); } } else if (curr->left == prev) { if (curr->right) { s.push(curr->right); } } else { s.pop(); } prev = curr; if (s.size() > max_depth) { max_depth = s.size(); } } return max_depth; } }; ~~~ ### 题解3 - 迭代(队列) 在使用了递归/后序遍历求解树最大深度之后,我们还可以直接从问题出发进行分析,树的最大深度即为广度优先搜索中的层数,故可以直接使用广度优先搜索求出最大深度。 ### C++ ~~~ /** * 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) { if (NULL == root) { return 0; } queue<TreeNode *> q; q.push(root); int max_depth = 0; while(!q.empty()) { int size = q.size(); for (int i = 0; i != size; ++i) { TreeNode *node = q.front(); q.pop(); if (node->left) { q.push(node->left); } if (node->right) { q.push(node->right); } } ++max_depth; } return max_depth; } }; ~~~ ### Java ~~~ /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: The root of binary tree. * @return: An integer. */ public int maxDepth(TreeNode root) { if (root == null) { return 0; } int depth = 0; Queue<TreeNode> q = new LinkedList<TreeNode>(); q.offer(root); while (!q.isEmpty()) { depth++; int qLen = q.size(); for (int i = 0; i < qLen; i++) { TreeNode node = q.poll(); if (node.left != null) q.offer(node.left); if (node.right != null) q.offer(node.right); } } return depth; } } ~~~ ### 源码分析 广度优先中队列的使用中,`qLen` 需要在for 循环遍历之前获得,因为它是一个变量。 ### 复杂度分析 最坏情况下空间复杂度为 O(n)O(n)O(n), 遍历每一个节点,时间复杂度为 O(n)O(n)O(n),