面试题：二叉树的前中后序遍历
二叉树的前中后遍历，其前中后，您可理解为指的是根结点所在的序。前序遍历：前序遍历的顺序为根-左-右中序遍历：中序遍历的顺序为左-根-右后序遍历：后序遍历的顺序为左-右-根
1.前序遍历
思路：利用栈后进先出的特性。1.根结点入栈；2.循环取栈顶元素、右子结点入栈、左子结点入栈。JAVA参考代码public class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int val) { this.val = val; }}public List<Integer> preorderTree(TreeNode root) { Stack<TreeNode> stack = new Stack<>(); List<Integer> preorder = new ArrayList<>(); if (root == null) { return preorder; } stack.push(root); while (!stack.empty()) { TreeNode node = stack.pop(); preorder.add(node.val); if (node.right != null) { stack.push(node.right); } if (node.left != null) { stack.push(node.left); } } return preorder;}
2.中序遍历
思路：利用栈后进先出的特性。1.根结点入栈；2.所有左子结点入栈；3.循环取出栈顶元素、判断右子结点是否为空：4.为空：取栈顶元素、若当前元素为栈顶元素右子树，弹出丢弃即可；（此时根结点已被上述访问取出）5.不为空：入栈、然后所有子节点入栈。JAVA参考代码public List<Integer> inorderTree(TreeNode root) { Stack<TreeNode> stack = new Stack<>(); List<Integer> inorder = new ArrayList<>(); while (root != null) { stack.push(root); root = root.left; } while (!stack.isEmpty()) { TreeNode node = stack.peek(); inorder.add(node.val); if (node.right == null) { node = stack.pop(); while (!stack.isEmpty() && stack.peek().right == node) { node = stack.pop(); } } else { node = node.right; while (node != null) { stack.push(node); node = node.left; } } } return inorder;}
3.后序遍历
思路：借助全局变量，递归左子结点、递归右子结点。JAVA参考代码List<Integer> result = new ArrayList<>();public List<Integer> postorderTree(TreeNode root) { TreeNode cur = root; if(cur == null) { return result; } if(cur.left != null) { postorderTree(cur.left); } if(cur.right != null) { postorderTree(cur.right); } result.add(cur.val); return result;}
