二叉树的遍历方式有广度还有深度方式
深度优先遍历:先往深走,遇到叶子节点再往回走。
广度优先遍历:一层一层的去遍历。
本文写的是深度优先遍历,分为前序,中序,后序遍历。这里前中后,其实指的就是中间节点的遍历顺序
- 二叉树的前序遍历
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {public List<Integer> preorderTraversal(TreeNode root) {List<Integer> result = new ArrayList<>();preHandle(root,result);return result;}private void preHandle(TreeNode root, List<Integer> result) {if(root== null){return;}result.add(root.val);preHandle(root.left,result);preHandle(root.right,result);}
}
- 二叉树的后序遍历
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {public List<Integer> postorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<>();postorder(root, res);return res;}private void postorder(TreeNode root, List<Integer> list) {if (root == null) {return;}postorder(root.left, list);postorder(root.right, list);list.add(root.val); // 注意这一句}
}
- 二叉树的中序遍历
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {public List<Integer> inorderTraversal(TreeNode root) {List<Integer> result = new ArrayList<>();//preHandle(treeNode,result);middleHandle(root,result);return result;}private void middleHandle(TreeNode root, List<Integer> result) {if(root== null){return;}middleHandle(root.left,result);result.add(root.val);middleHandle(root.right,result);}
}
迭代遍历
递归的实现就是:每一次递归调用都会把函数的局部变量、参数值和返回地址等压入调用栈中,然后递归返回的时候,从栈顶弹出上一次递归的各项参数,所以这就是递归为什么可以返回上一层位置的原因。
// 前序遍历顺序:中-左-右,入栈顺序:中-右-左public List<Integer> preorderTraversal(TreeNode root) {List<Integer> result = new ArrayList<>();if (root == null){return result;}Stack<TreeNode> stack = new Stack<>();stack.push(root);while (!stack.isEmpty()){TreeNode node = stack.pop();result.add(node.val);if (node.right != null){stack.push(node.right);}if (node.left != null){stack.push(node.left);}}return result;}// 中序遍历顺序: 左-中-右 入栈顺序: 左-右public List<Integer> inorderTraversal(TreeNode root) {List<Integer> result = new ArrayList<>();if (root == null){return result;}Stack<TreeNode> stack = new Stack<>();TreeNode cur = root;while (cur != null || !stack.isEmpty()){if (cur != null){stack.push(cur);cur = cur.left;}else{cur = stack.pop();result.add(cur.val);cur = cur.right;}}return result;}// 后序遍历顺序 左-右-中 入栈顺序:中-左-右 出栈顺序:中-右-左, 最后翻转结果public List<Integer> postorderTraversal(TreeNode root) {List<Integer> result = new ArrayList<>();if (root == null){return result;}Stack<TreeNode> stack = new Stack<>();stack.push(root);while (!stack.isEmpty()){TreeNode node = stack.pop();result.add(node.val);if (node.left != null){stack.push(node.left);}if (node.right != null){stack.push(node.right);}}Collections.reverse(result);return result;}