Leetcode 112. Path Sum

方法1: recursion，其实也是dfs。时间复杂n，空间复杂logn。

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        TreeNode left = root.left;
        TreeNode right = root.right;
        if(left == null && right == null && root.val == sum) return true;
        return hasPathSum(left, sum-root.val) || hasPathSum(right, sum-root.val);
    }
}

方法2: dfs using stack/iteration。这边依旧是用了两个stack，一个用来装treenode，另一个用来存储curr_sum。时间复杂n，空间复杂logn。

public boolean hasPathSum(TreeNode root, int sum) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        Stack<Integer> sums = new Stack<Integer>();

        stack.push(root);
        sums.push(sum);
        while(!stack.isEmpty()&&(root!=null)){
            int value = sums.pop();
            TreeNode top = stack.pop();
            
            if(top.left==null&&top.right==null&&top.val==value){
                return true;
            }
            if(top.right!=null){
                stack.push(top.right);
                sums.push(value-top.val);
            }
            if(top.left!=null){
                stack.push(top.left);
                sums.push(value-top.val);
            }

        }
        return false;
    }

总结：

• 这题dfs其实是preorder traversal。root->left->right。