LeetCode 145 Binary Tree Postorder Traversal

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?

方法一 递归

public List<Integer> postorderTraversal2(TreeNode root) {
List<Integer> result = new ArrayList<>();
postorderTraversal(root, result);
return result;
}

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

方法二： 后序遍历是 左 右 根 的顺序，所以我们现在进行 根 右 左 的顺序，最后倒序过来 就是想要的左右根的顺序了

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

}
Collections.reverse(list);
return list;
}

方法三 使用 LinkedList 

public List<Integer> postorderTraversal6(TreeNode root) {
LinkedList<Integer> list = new LinkedList<>();
if (root == null) return list;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode temp = stack.pop();
list.addFirst(temp.val);
if (temp.left != null) stack.push(temp.left);
if (temp.right != null) stack.push(temp.right);
}
return list;
}

方法四：如果node的孩子已经入栈，那么就把node的对应孩子置为空，防止后续访问此节点时，不知是该入栈其孩子，还是出栈此节点

public List<Integer> postorderTraversal3(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if (root == null) return res;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode temp = stack.peek();
if (temp.left == null && temp.right == null) {
TreeNode pop = stack.pop();
res.add(pop.val);
} else {
if (temp.right != null) {
stack.push(temp.right);
temp.right = null;//这步很巧妙
}
if (temp.left != null) {
stack.push(temp.left);
temp.left = null;
}
}
}
return res;
}

方法四：不借助于栈http://www.cnblogs.com/yrbbest/p/4489571.html

public List<Integer> postorderTraversal4(TreeNode root) {
List<Integer> res = new LinkedList<>();
TreeNode node = root, succ;
while (node != null) {
if (node.right == null) {
res.add(0, node.val);
node = node.left;
} else {
succ = node.right;
while (succ.left != null && succ.left != node) {
succ = succ.left;
}
if (succ.left == null) {
succ.left = node;
res.add(0, node.val);
node = node.right;
} else {
succ.left = null;
node = node.left;
}
}
}
return res;
}


http://blog.csdn.net/crazy1235/article/details/51494797