LeetCode- 树的前序、中序、后序遍历（递归与迭代）

144. Binary Tree Preorder Traversal

前序遍历

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

递归

public List<Integer> preorderTraversal(TreeNode root) {List<Integer> list = new ArrayList<>();if (root == null)return list;preorder(root, list);return list;}private void preorder(TreeNode root, List<Integer> list) {if (root != null) {list.add(root.val);preorder(root.left, list);preorder(root.right, list);}

迭代

public List<Integer> preorderTraversal(TreeNode root) {List<Integer> list = new ArrayList<>();if (root == null)return list;Stack<TreeNode> stack = new Stack<>();stack.push(root);while (!stack.isEmpty()) {//出栈TreeNode node=stack.pop();list.add(node.val);//访问右子树if(node.right!=null)stack.push(node.right);//左子树if(node.left!=null)stack.push(node.left);}return list;}

94. Binary Tree Inorder Traversal

中序遍历

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

   1    \     2    /   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

递归

public List<Integer> inorderTraversal(TreeNode root) {List<Integer> list = new ArrayList<>();if (root == null)return list;inorder(root, list);return list;}private void inorder(TreeNode root, List<Integer> list) {if (root != null) {inorder(root.left, list);list.add(root.val);inorder(root.right, list);}}

迭代

public List<Integer> inorderTraversal(TreeNode root) {List<Integer> list = new ArrayList<>();if (root == null)return list;Stack<TreeNode> stack = new Stack<>();while (root != null || !stack.isEmpty()) {if (root != null) {stack.push(root);root = root.left;} else {root = stack.pop();list.add(root.val);root = root.right;}}return list;}

145. Binary Tree Postorder Traversal

后序遍历

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

递归

public List<Integer> postorderTraversal(TreeNode root) {List<Integer> list = new ArrayList<>();if (root == null)return list;postorder(root, list);return list;}private void postorder(TreeNode root, List<Integer> list) {if (root != null) {postorder(root.left, list);postorder(root.right, list);list.add(root.val);}}

迭代

关键点是找一个flag存储刚刚访问过的结点

public List<Integer> postorderTraversal(TreeNode root) {List<Integer> list = new ArrayList<>();if (root == null)return list;TreeNode p = root;Stack<TreeNode> stack = new Stack<>();// 辅助指针用于标记最近访问的结点，// 标记是从左子树返回还是右子树返回TreeNode flag = null;while (p != null || !stack.isEmpty()) {if (p != null) {// 走到最左边stack.push(p);p = p.left;}// 向右else {// 去栈顶结点p = stack.peek();// 若右子树存在且未被访问过if (p.right != flag && p.right != null) {// 转向右p = p.right;stack.push(p);// 在走到最左p = p.left;} else {// 访问p = stack.pop();list.add(p.val);// 记录最近访问的结点flag = p;// 结点访问完毕后重置指针pp = null;}}}return list;}