lintcode 访问路径
http://www.lintcode.com/zh-cn/problem/binary-tree-postorder-traversal/
描述
给出一棵二叉树，返回其节点值的后序遍历。
样例
给出一棵二叉树 {1,#,2,3},
返回 [3,2,1]
何为二叉树后序遍历
顺序：左 -> 右 -> 根
获取到一个节点后，将其暂存，遍历完左右子树后，再输出该节点的值。
Java代码实现
非递归实现
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| /**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: Postorder in ArrayList which contains node values.
*/
public ArrayList<Integer> postorderTraversal(TreeNode root) {
// write your code here
ArrayList<Integer> result = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode node = root;
TreeNode lastVisit = root;
while (node != null || !stack.isEmpty()) {
while (node != null) {
stack.push(node);
node = node.left;
}
node = stack.peek();
if (node.right == null || node.right == lastVisit) {
result.add(node.val);
stack.pop();
lastVisit = node;
node = null;
} else {
node = node.right;
}
}
return result;
}
}
|
递归实现
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| /**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: Postorder in ArrayList which contains node values.
*/
public ArrayList<Integer> postorderTraversal(TreeNode root) {
// write your code here
ArrayList<Integer> result = new ArrayList<Integer>();
if (root == null) {
return result;
}
result.addAll(postorderTraversal(root.left));
result.addAll(postorderTraversal(root.right));
result.add(root.val);
return result;
}
}
|
Python代码实现
非递归实现
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| """
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: The root of binary tree.
@return: Postorder in ArrayList which contains node values.
"""
def postorderTraversal(self, root):
# write your code here
result = []
stack = []
node = root
last_visit = root
while node or stack:
while node:
stack.append(node)
node = node.left
node = stack[-1]
if node.right is None or node.right == last_visit:
result.append(node.val)
stack.pop()
last_visit = node
node = None
else:
node = node.right
return result
|
递归实现
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| """
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: The root of binary tree.
@return: Postorder in ArrayList which contains node values.
"""
def postorderTraversal(self, root):
# write your code here
self.result = []
self.traversal(root)
return self.result
def traversal(self, root):
if root is None:
return
self.traversal(root.left)
self.traversal(root.right)
self.result.append(root.val)
|