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代码随想录算法训练营第十四天｜二叉树遍历

Posted on Edited on

今天学习了如何使用递归法对二叉树进行前序、中序和后序遍历

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// pre-order
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        
        List<Integer> ans = new LinkedList<>();
        if(root != null) {
            preorder(ans, root);
            return ans;
        }

        return ans;

    }

    private void preorder ( List<Integer> ans, TreeNode next) {

        // as this is pre-order scan
        // no matter whether this node has children or not
        // add its value in the ans first
        TreeNode node = next;
        ans.add(node.val);

        // if there is no child, just return
        if ( node.left == null && node.right == null) {
            return;
            // else let's call this method recursively on its children
        }else{
            if(node.left != null) {
                preorder(ans, node.left);
            }
    
            if(node.right != null) {
                preorder(ans, node.right);
            }
        }
    }
}

// in-order
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
            List<Integer> ans = new LinkedList<>();

            if (root != null){
                inorder(ans, root);
            }

            return ans;
    }

    private void inorder (List<Integer> ans, TreeNode next) {
        TreeNode node = next;
        if (node.left == null){
            ans.add(node.val);

            if (node.right != null) {
                inorder(ans, node.right);
            }else{
                return;
            }
        }else{
            inorder(ans, node.left);
            ans.add(node.val);
            if (node.right != null) {
                inorder(ans, node.right);
            }
        }
    }
}

// post-order
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> ans = new LinkedList<>();

        if (root != null){
            postorder(ans, root);
        }

        return ans;
    }

    private void postorder(List<Integer> ans, TreeNode next) {
        TreeNode node = next;

        if (node.left == null && node.right == null) {
            ans.add(node.val);
            return;
        }else if (node.left != null && node.right != null){
            postorder(ans, node.left);
            postorder(ans, node.right);
            ans.add(node.val);
        }else {
            if ( node.left != null) {
                postorder(ans, node.left);
                ans.add(node.val);
            }

            if ( node.right != null) {
                postorder(ans, node.right);
                ans.add(node.val);
            }
        }
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */

一开始写的不够简洁，修改了一下递归出口，修改后：

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
  
// pre-order
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ans = new LinkedList<>();
        preorder(ans, root);
        return ans;

    }

    private void preorder ( List<Integer> ans, TreeNode cur) {
        // recursion exitrance condition
        if ( cur == null) {
            return;
        }
        // pre-order
        ans.add(cur.val);
        preorder(ans, cur.left);
        preorder(ans, cur.right);
    }
}

// in-order
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> ans = new LinkedList<>();
        inorder(ans, root);
        return ans;
    }

    private void inorder (List<Integer> ans, TreeNode cur) {
        // recursion exitrance condition
        if ( cur == null) {
            return;
        }

        // inorder 
        inorder(ans, cur.left);
        ans.add(cur.val);
        inorder(ans, cur.right);
    }
}

// post-order
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> ans = new LinkedList<>();
        postorder(ans, root);
        return ans;
    }

    private void postorder(List<Integer> ans, TreeNode cur) {

        // recursion exitrance condition
        if (cur == null) {
            return;
        }

        // post-order
        postorder(ans, cur.left);
        postorder(ans, cur.right);
        ans.add(cur.val);
    }
}

除了使用递归方法，我们还可以使用一个栈结构来模拟递归行为：

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// pre-order
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ans = new LinkedList<>();

        if (root == null) {
            return ans;
        }

        Stack<TreeNode> stack = new Stack<>();
        TreeNode node;
        stack.push(root);
      
        while (!stack.isEmpty()) {
            node = stack.pop();
            ans.add(node.val);

            if (node.right != null) {
                stack.push(node.right);
            }

            if (node.left != null) {
                stack.push(node.left);
            }

            
        }

        return ans;

    }
}

// in-order
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> ans = new LinkedList<>();

        if (root == null) {
            return ans;
        }

        Stack<TreeNode> visited = new Stack<>();
        TreeNode node = root;
        while (node != null || !visited.isEmpty()) {

            // go along the left side until touching the leftmost leaf node
            if (node != null) {
                visited.push(node);
                node = node.left;
            }else {

                // pop the last element we visited, records its value and 
                // point to its right node
                node = visited.pop();
                ans.add(node.val);
                node = node.right;
            }

        }
        return ans;
    }

}

// post-order
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> ans = new LinkedList<>();

        if (root == null) {
            return ans;
        }

        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;

        // post order left right root, reverse to root, right, left in stack
        // see what? root, right, left, similar to pre-order but get child nodes swapped
        // thus, it is easy to think how about let's slightly change the code for pre-order
        stack.push(node);
        while (!stack.isEmpty()) {
            ans.add(node.val);

            if (node.left != null) {
                stack.push(node.left);
            }

            if (node.right != null) {
                stack.push(node.right);
            }

            node = stack.pop();
        }

        // now we get the values but in root right left, we need to reverse the ans list
        for (int i = 0, j = ans.size() - 1, temp = 0; i < j; i++, j--) {
            temp = ans.get(i);
            ans.set(i, ans.get(j));
            ans.set(j, temp);
        }

        return ans;
    }
}