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難度：容易
要求：
給出一棵二叉樹,返回其后序遍歷
樣例給出一棵二叉樹
{1,#,2,3}
1
\
2
/
3
返回 [3,2,1]
思路：非遞歸
/**
* 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 {
public ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> result = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode prev = null; // previously traversed node
TreeNode curr = root;
if (root == null) {
return result;
}
stack.push(root);
while (!stack.empty()) {
curr = stack.peek();
if (prev == null || prev.left == curr || prev.right == curr) { // traverse down the tree
if (curr.left != null) {
stack.push(curr.left);
} else if (curr.right != null) {
stack.push(curr.right);
}
} else if (curr.left == prev) { // traverse up the tree from the left
if (curr.right != null) {
stack.push(curr.right);
}
} else { // traverse up the tree from the right
result.add(curr.val);
stack.pop();
}
prev = curr;
}
return result;
}
}
思路：遞歸
/**
* 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 {
public ArrayList<Integer> postorderTraversal(TreeNode root) {
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;
}
}
