[LintCode/LeetCode] Binary Tree InOrder Traversal

tomorrowwu 發(fā)布于2019-08-14 15:39 / 2793人閱讀

摘要：遞歸法不說(shuō)了，棧迭代的函數(shù)是利用的原理，從根節(jié)點(diǎn)到最底層的左子樹，依次入堆棧。然后將出的結(jié)點(diǎn)值存入數(shù)組，并對(duì)出的結(jié)點(diǎn)的右子樹用函數(shù)繼續(xù)迭代。

Problem

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

Example

Given binary tree {1,#,2,3},

return [1,3,2].

Challenge

Can you do it without recursion?

Note

遞歸法不說(shuō)了，棧迭代的helper函數(shù)是利用stack的LIFO原理，從根節(jié)點(diǎn)到最底層的左子樹，依次push入堆棧。然后將pop出的結(jié)點(diǎn)值存入數(shù)組，并對(duì)pop出的結(jié)點(diǎn)的右子樹用helper函數(shù)繼續(xù)迭代。

Solution

Recursion

public class Solution {
    public ArrayList inorderTraversal(TreeNode root) {
        ArrayList res = new ArrayList();
        if (root == null) return res;
        helper(root, res);
        return res;
    }
    public void helper(TreeNode root, ArrayList res) {
        if (root.left != null) helper(root.left, res);
        res.add(root.val);
        if (root.right != null) helper(root.right, res);
    }
}

Stack Iteration

public class Solution {
    public ArrayList inorderTraversal(TreeNode root) {
        ArrayList res = new ArrayList();
        Stack stack = new Stack();
        if (root == null) return res;
        helper(stack, root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if (cur.right != null) helper(stack, cur.right);
        }
        return res;
    }
    public void helper(Stack stack, TreeNode root) {
        stack.push(root);
        while (root.left != null) {
            root = root.left;
            stack.push(root);
        }
    }
}

Another Stack Iteration

public class Solution {
    public ArrayList inorderTraversal(TreeNode root) {
        ArrayList res = new ArrayList<>();
        Stack stack = new Stack<>();
        TreeNode cur = root;
        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.add(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            res.add(cur.val);
            cur = cur.right;
        }
        return res;
    }
}