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.
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its depth = 3.
题意:求二叉树最大深度。
思路:
方法1:递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if(root==null) return 0;
return 1+Math.max(maxDepth(root.left),maxDepth(root.right));
}
}
方法二:迭代 利用栈
考虑节点栈不为空和当前节点不为空
public static int TreeDepth(TreeNode root) {
TreeNode node = root;
Stack<TreeNode> nodeStack = new Stack<TreeNode>();
Stack<Integer> depthStack = new Stack<Integer>();
int depth = 1;
int max = 0;
while(node != null || nodeStack.size()>0){
if(node!= null){
nodeStack.push(node);
depthStack.push(depth);
node = node.left;
depth++;
}else{
node = nodeStack.pop();
depth = depthStack.pop();
if(depth > max){
max = depth;
}
node = node.right;
depth++;
}
}
return max;
}
方法三:
BFS
遍历每一层放进队列
class Solution {
public int maxDepth(TreeNode root) {
if (root == null) return 0;
int depth = 0;
Queue<TreeNode> nodes = new LinkedList<>();
nodes.offer(root);
while (!nodes.isEmpty()) {
int size = nodes.size();
depth++;
while (size-- > 0) {
TreeNode node = nodes.poll();
if (node.left != null) nodes.offer(node.left);
if (node.right != null) nodes.offer(node.right);
}
}
return depth;
}
}
测试用例:
public class Solution {
public static void main(String[] args){
TreeNode node0 = new TreeNode(0);
TreeNode node1 = new TreeNode(1);
TreeNode node2 = new TreeNode(2);
TreeNode node3 = new TreeNode(3);
TreeNode node4 = new TreeNode(4);
node0.left = node1;
node0.right = node2;
node1.left = node3;
node3.left = node4;
System.out.println(TreeDepth(node0));
}
public static int TreeDepth(TreeNode root) {
return root!=null ? 1+Math.max(TreeDepth(root.left),TreeDepth(root.right)):0;
}
}
class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}