104.Maximum-Depth-of-Binary-Tree

104. Maximum Depth of Binary Tree

题目地址

https://leetcode.com/problems/maximum-depth-of-binary-tree

http://www.lintcode.com/problem/maximum-depth-of-binary-tree/

http://www.jiuzhang.com/solutions/maximum-depth-of-binary-tree/

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.

Note: A leaf is a node with no children.

Example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7
return its depth = 3.

代码

Approach 1: Divide Conquer (Recursive)

Intuition By definition, the maximum depth of a binary tree is the maximum number of steps to reach a leaf node from the root node.

Complexity analysis

• Time complexity : we visit each node exactly once, thus the time complexity is O(N), where N is the number of nodes.

• Space complexity : in the worst case, the tree is completely unbalanced, e.g. each node has only left child node, the recursion call would occur N times (the height of the tree), therefore the storage to keep the call stack would be O(N). But in the best case (the tree is completely balanced), the height of the tree would be log(N). Therefore, the space complexity in this case would be O(log(N)).

public class Solution {
    public int maxDepth(TreeNode root) {
    if (root == null) return 0;
  }

  int left = maxDepth(root.left);
  int right = maxDepth(root.right);
  return Math.max(left, right) + 1;
}

Approach 2: Traverse

class Solution {
  private int depth;

  public int maxDepth(TreeNode root) {
    depth = 0;
    helper(root, 1);

    return depth;
  }

  private void helper(TreeNode node, int curtDepth) {
    if (node == null) return;

    if (curtDepth > depth) {
      depth = curtDepth;
    }

    helper(node.left, curtDepth + 1);
    helper(node.right, curtDepth + 1);
  }
}

Approach 3: Iteration

class Solution {
  public int maxDepth(TreeNode root) {
    LinkedList<TreeNode> stack = new LinkedList<>();
    LinkedList<TreeNode> depths = new LinkedList<>();
    if (root == null)    return 0;

    stack.add(root);
    depths.add(1);

    int depth = 0;
    int current_depth = 0;
    while (!stack.isEmpty()) {
      root = stack.pollLast();
      current_depth = depths.pollLast();
      if (root != null) {
        depth = Math.max(depth, current_depth);
        stack.add(root.left);
        stack.add(root.right);
        depths.add(current_depth + 1);
        depths.add(current_depth + 1);
      }
    }

    return depth;
  }
}