二叉树的中序遍历
按照访问左子树——根节点——右子树的方式遍历这棵树.
定义 inorder(root) 表示当前遍历到 \textit{root}root 节点的答案,那么按照定义,我们只要递归调用 inorder(root.left) 来遍历 \textit{root}root 节点的左子树,然后将 \textit{root}root 节点的值加入答案,再递归调用inorder(root.right) 来遍历 \textit{root}root 节点的右子树即可,递归终止的条件为碰到空节点。
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
inorder(root, res);
return res;
}
public void inorder(TreeNode root, List<Integer> res) {
if (root == null) {
return;
}
inorder(root.left, res);
res.add(root.val);
inorder(root.right, res);
}
}
附加前序遍历代码
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
preorder(root,res);
return res;
}
public void preorder(TreeNode root, List<Integer> res){
if(root == null){
return ;
}
res.add(root.val);
preorder(root.left,res);
preorder(root.right,res);
}
}
附加后序遍历代码
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
postorder(root, res);
return res;
}
public void postorder(TreeNode root, List<Integer> res) {
if (root == null) {
return;
}
postorder(root.left, res);
postorder(root.right, res);
res.add(root.val);
}
}
相同的树
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;
else
return isSameTree(p.left,q.left)&& isSameTree(p.right,q.right);
}
}
对称二叉树
class Solution {
public boolean isSymmetric(TreeNode root) {
return check(root,root);
}
public boolean check(TreeNode p,TreeNode q){
if(p==null&&q==null)
return true;
else if(p==null || q==null)
return false;
else
return p.val==q.val && check(p.left,q.right) && check(p.right,q.left);
}
}
平衡二叉树
class Solution {
public boolean isBalanced(TreeNode root) {
if (root == null) {
return true;
} else {
return Math.abs(height(root.left) - height(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right);
}
}
public int height(TreeNode root) {
if (root == null) {
return 0;
} else {
return Math.max(height(root.left), height(root.right)) + 1;
}
}
}
二叉树的最大深度
class Solution {
public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
} else {
int leftHeight = maxDepth(root.left);
int rightHeight = maxDepth(root.right);
return Math.max(leftHeight, rightHeight) + 1;
}
}
}
附加最小深度
class Solution {
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
if (root.left == null && root.right == null) {
return 1;
}
int min_depth = Integer.MAX_VALUE;
if (root.left != null) {
min_depth = Math.min(minDepth(root.left), min_depth);
}
if (root.right != null) {
min_depth = Math.min(minDepth(root.right), min_depth);
}
return min_depth + 1;
}
}
路径总和
class Solution {
public boolean hasPathSum(TreeNode root, int targetSum) {
if(root==null)
return false;
if(root.left==null&&root.right==null)
return targetSum==root.val;
return hasPathSum(root.left,targetSum-root.val)||hasPathSum(root.right,targetSum-root.val);
}
}