/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int maxDepth(TreeNode root) {
if (root == null)
return 0;
int leftHeight = maxDepth(root.left);
int rightHeight = maxDepth(root.right);
return Math.max(leftHeight, rightHeight) + 1;
}
}
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
Firstly, I use a more brute force method. I did not think a better way to deal with the one of the p, q is null issue, so I write a lot of if else.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null)
return true;
else if (p == null && q != null)
return false;
else if (p != null && q == null)
return false;
else if (p.val != q.val)
return false;
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}
Later I find a better way to deal with this if else condition for p, q is null issue to make the code shorter and more clear.
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null)
return true;
else if (p==null || q==null)
return false;
else if (p.val != q.val)
return false;
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}
原文：http://www.cnblogs.com/timoBlog/p/4635193.html