二叉树的遍历

• 前序遍历 preorder：根左右
• 中序遍历 inorder：左根右
• 后序遍历 postorder：左右根
  每一种遍历都有递归和循环两种方法，其中递归比循环简单得多。
  三种遍历递归方法的区别就是递归函数中递归左孩子、右孩子与当前节点的顺序。
  而循环则需借用栈来实现。

前序遍历

递归

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> preorderTraversal(TreeNode root) {        List<Integer> result = new ArrayList<Integer>();        helper(result,root);        return result;    }    public void helper(List<Integer> list,TreeNode node){        if(node == null) return;        //根左右        list.add(node.val);        helper(list,node.left);        helper(list,node.right);    }}

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循环

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> preorderTraversal(TreeNode root) {        List<Integer> result = new ArrayList<Integer>();        if(root == null) return result;        Stack<TreeNode> stack = new Stack<TreeNode>();        stack.push(root);        while(!stack.isEmpty()){            TreeNode node = stack.pop();            result.add(node.val);            if(node.right!=null){                stack.push(node.right);            }            if(node.left!=null){                stack.push(node.left);            }        }        return result;    }}

中序遍历

递归

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> inorderTraversal(TreeNode root) {        List<Integer> result = new ArrayList<Integer>();        helper(result,root);        return result;    }    public void helper(List<Integer> list,TreeNode node){        if(node == null) return;        //左根右        helper(list,node.left);        list.add(node.val);        helper(list,node.right);    }}

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循环

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> inorderTraversal(TreeNode root) {        List<Integer> result = new ArrayList<Integer>();        Stack<TreeNode> stack = new Stack<TreeNode>();        TreeNode node = root;        while(!stack.isEmpty() || node != null){            while(node != null){                stack.push(node);                node = node.left;            }            node = stack.pop();            result.add(node.val);            node = node.right;        }        return result;    }}

后序遍历

递归

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> postorderTraversal(TreeNode root) {        List<Integer> result = new ArrayList<Integer>();        helper(result,root);        return result;    }    public void helper(List<Integer> list,TreeNode node){        if(node == null) return;        //左右根        helper(list,node.left);        helper(list,node.right);        list.add(node.val);    }}

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循环

用prev表示上一次遍历到的节点，curr是当前遍历的节点，需根据二者的关系来确定接下来的遍历方向。

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> postorderTraversal(TreeNode root) {        List<Integer> result = new ArrayList<Integer>();        Stack<TreeNode> stack = new Stack<TreeNode>();        if(root == null) return;        TreeNode prev = null;        TreeNode curr = root;        stack.push(root);        while(!stack.isEmpty()){            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;    }}